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sửa đổi
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giúp vs
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Từ giả thiết ta cóa(c2+b2)=c(a2+b2)⇔ac2+ab2=ca2+cb2⇔(b2−ac)(a−c)=0" role="presentation" style="display: inline-block; line-height: 0; font-size: 16.38px; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; padding-top: 1px; padding-bottom: 1px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">a(c2+b2)=c(a2+b2)⇔ac2+ab2=ca2+cb2⇔(b2−ac)(a−c)=0a(c2+b2)=c(a2+b2)⇔ac2+ab2=ca2+cb2⇔(b2−ac)(a−c)=0Mà a−c≠0⇒b2=ac" role="presentation" style="display: inline-block; line-height: 0; font-size: 16.38px; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; padding-top: 1px; padding-bottom: 1px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">a−c≠0⇒b2=aca−c≠0⇒b2=acMà a,b,c" role="presentation" style="display: inline-block; line-height: 0; font-size: 16.38px; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; padding-top: 1px; padding-bottom: 1px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">a,b,ca,b,c là các số nguyên tố⇒a=b=c" role="presentation" style="display: inline-block; line-height: 0; font-size: 16.38px; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; padding-top: 1px; padding-bottom: 1px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">⇒a=b=c⇒a=b=c⇒a2+b2+c2" role="presentation" style="display: inline-block; line-height: 0; font-size: 16.38px; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; padding-top: 1px; padding-bottom: 1px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">⇒a2+b2+c2⇒a2+b2+c2 không thể là số nguyên tố
Từ GT ta có :a(c2+b2)=c(a2+b2)⇔ac2+ab2=ca2+cb2⇔(b2−ac)(a−c)=0" role="presentation" style="display: inline-block; line-height: 0; font-size: 16.38px; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; padding-top: 1px; padding-bottom: 1px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">a(c2+b2)=c(a2+b2)⇔ac2+ab2=ca2+cb2⇔(b2−ac)(a−c)=0a(c2+b2)=c(a2+b2)⇔ac2+ab2=ca2+cb2⇔(b2−ac)(a−c)=0Mà a−c≠0⇒b2=ac" role="presentation" style="display: inline-block; line-height: 0; font-size: 16.38px; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; padding-top: 1px; padding-bottom: 1px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">a−c≠0⇒b2=aca−c≠0⇒b2=acMà a,b,c" role="presentation" style="display: inline-block; line-height: 0; font-size: 16.38px; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; padding-top: 1px; padding-bottom: 1px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">a,b,ca,b,c là các số nguyên tố⇒a=b=c" role="presentation" style="display: inline-block; line-height: 0; font-size: 16.38px; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; padding-top: 1px; padding-bottom: 1px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">⇒a=b=c⇒a=b=c(vô lí)⇒a2+b2+c2" role="presentation" style="display: inline-block; line-height: 0; font-size: 16.38px; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; padding-top: 1px; padding-bottom: 1px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">⇒a2+b2+c2⇒a2+b2+c2 không thể là số nguyên tố
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sửa đổi
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giúp vs
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Từ giả thiết ta cóa(c2+b2)=c(a2+b2)⇔ac2+ab2=ca2+cb2⇔(b2−ac)(a−c)=0" role="presentation" style="display: inline-block; line-height: 0; font-size: 16.38px; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; padding-top: 1px; padding-bottom: 1px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">a(c2+b2)=c(a2+b2)⇔ac2+ab2=ca2+cb2⇔(b2−ac)(a−c)=0a(c2+b2)=c(a2+b2)⇔ac2+ab2=ca2+cb2⇔(b2−ac)(a−c)=0Mà a−c≠0⇒b2=ac" role="presentation" style="display: inline-block; line-height: 0; font-size: 16.38px; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; padding-top: 1px; padding-bottom: 1px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">a−c≠0⇒b2=aca−c≠0⇒b2=acMà a,b,c" role="presentation" style="display: inline-block; line-height: 0; font-size: 16.38px; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; padding-top: 1px; padding-bottom: 1px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">a,b,ca,b,c là các số nguyên tố⇒a=b=c" role="presentation" style="display: inline-block; line-height: 0; font-size: 16.38px; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; padding-top: 1px; padding-bottom: 1px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">⇒a=b=c(vô lí)⇒a=b=c⇒a2+b2+c2" role="presentation" style="display: inline-block; line-height: 0; font-size: 16.38px; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; padding-top: 1px; padding-bottom: 1px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">⇒a2+b2+c2⇒a2+b2+c2 không thể là số nguyên tố
Từ giả thiết ta cóa(c2+b2)=c(a2+b2)⇔ac2+ab2=ca2+cb2⇔(b2−ac)(a−c)=0" role="presentation" style="display: inline-block; line-height: 0; font-size: 16.38px; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; padding-top: 1px; padding-bottom: 1px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">a(c2+b2)=c(a2+b2)⇔ac2+ab2=ca2+cb2⇔(b2−ac)(a−c)=0a(c2+b2)=c(a2+b2)⇔ac2+ab2=ca2+cb2⇔(b2−ac)(a−c)=0Mà a−c≠0⇒b2=ac" role="presentation" style="display: inline-block; line-height: 0; font-size: 16.38px; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; padding-top: 1px; padding-bottom: 1px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">a−c≠0⇒b2=aca−c≠0⇒b2=acMà a,b,c" role="presentation" style="display: inline-block; line-height: 0; font-size: 16.38px; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; padding-top: 1px; padding-bottom: 1px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">a,b,ca,b,c là các số nguyên tố⇒a=b=c" role="presentation" style="display: inline-block; line-height: 0; font-size: 16.38px; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; padding-top: 1px; padding-bottom: 1px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">⇒a=b=c⇒a=b=c⇒a2+b2+c2" role="presentation" style="display: inline-block; line-height: 0; font-size: 16.38px; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; padding-top: 1px; padding-bottom: 1px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">⇒a2+b2+c2⇒a2+b2+c2 không thể là số nguyên tố
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sửa đổi
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giúp vs
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Từ giả thiết ta cóa(c2+b2)=c(a2+b2)⇔ac2+ab2=ca2+cb2⇔(b2−ac)(a−c)=0" role="presentation" style="display: inline-block; line-height: 0; font-size: 16.38px; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; padding-top: 1px; padding-bottom: 1px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">a(c2+b2)=c(a2+b2)⇔ac2+ab2=ca2+cb2⇔(b2−ac)(a−c)=0a(c2+b2)=c(a2+b2)⇔ac2+ab2=ca2+cb2⇔(b2−ac)(a−c)=0Mà a−c≠0⇒b2=ac" role="presentation" style="display: inline-block; line-height: 0; font-size: 16.38px; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; padding-top: 1px; padding-bottom: 1px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">a−c≠0⇒b2=aca−c≠0⇒b2=acMà a,b,c" role="presentation" style="display: inline-block; line-height: 0; font-size: 16.38px; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; padding-top: 1px; padding-bottom: 1px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">a,b,ca,b,c là các số nguyên tố⇒a=b=c" role="presentation" style="display: inline-block; line-height: 0; font-size: 16.38px; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; padding-top: 1px; padding-bottom: 1px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">⇒a=b=c⇒a=b=c⇒a2+b2+c2" role="presentation" style="display: inline-block; line-height: 0; font-size: 16.38px; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; padding-top: 1px; padding-bottom: 1px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">⇒a2+b2+c2⇒a2+b2+c2 không thể là số nguyên tố
Từ giả thiết ta cóa(c2+b2)=c(a2+b2)⇔ac2+ab2=ca2+cb2⇔(b2−ac)(a−c)=0" role="presentation" style="display: inline-block; line-height: 0; font-size: 16.38px; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; padding-top: 1px; padding-bottom: 1px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">a(c2+b2)=c(a2+b2)⇔ac2+ab2=ca2+cb2⇔(b2−ac)(a−c)=0a(c2+b2)=c(a2+b2)⇔ac2+ab2=ca2+cb2⇔(b2−ac)(a−c)=0Mà a−c≠0⇒b2=ac" role="presentation" style="display: inline-block; line-height: 0; font-size: 16.38px; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; padding-top: 1px; padding-bottom: 1px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">a−c≠0⇒b2=aca−c≠0⇒b2=acMà a,b,c" role="presentation" style="display: inline-block; line-height: 0; font-size: 16.38px; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; padding-top: 1px; padding-bottom: 1px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">a,b,ca,b,c là các số nguyên tố⇒a=b=c" role="presentation" style="display: inline-block; line-height: 0; font-size: 16.38px; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; padding-top: 1px; padding-bottom: 1px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">⇒a=b=c(vô lí)⇒a=b=c⇒a2+b2+c2" role="presentation" style="display: inline-block; line-height: 0; font-size: 16.38px; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; padding-top: 1px; padding-bottom: 1px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">⇒a2+b2+c2⇒a2+b2+c2 không thể là số nguyên tố
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bình luận
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giúp vs
đúng tick V nha bạn
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bình luận
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giúp vs
bài này không khó bạn chỉ để ý một chút sẽ ra
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giải đáp
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giúp vs
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Từ GT ta có :
a(c2+b2)=c(a2+b2)⇔ac2+ab2=ca2+cb2⇔(b2−ac)(a−c)=0
Mà a−c≠0⇒b2=ac
Mà a,b,c là các số nguyên tố
⇒a=b=c(vô lí)
⇒a2+b2+c2 không thể là số nguyên tố
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giải đáp
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Mn ủng hộ , tạm 10 câu đã hì hì
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Câu 9 $B=-x+\sqrt{x-2}+2\sqrt{x+1}+2014=\sqrt{\left ( x-2 \right )1}-x+2\sqrt{x+1}+2014\leq \frac{x-2+1}{2}-x+2\sqrt{x+1}+2014=\frac{-x}{2}+2\sqrt{x+1}+2014-\frac{1}{2}=\frac{-1}{2}\left ( x +1-4\sqrt{x+1}+4\right )+2016=\frac{-1}{2}\left ( \sqrt{x+1} -2\right )^{2}+2016\leq 2016$ Dấu"=" xảy ra khi \begin{cases}x-2=1 \\ \left ( \sqrt{x+1}-2\right )^{2}=0 \end{cases}$\Leftrightarrow x=3$ Vậy max B=2016 khi và chỉ khi x=3 |
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sửa đổi
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giúp mình với !!!!!!!!!!!!!!!!
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Pt đã cho $ (a+b)^{3}+c^{3}-3ab(a+b)-3abc=0$$\Leftrightarrow \left ( a+b+c \right )\left ( a^{2}+b^{2}+c^{2} -ab-bc-ca\right )=0$$\Leftrightarrow a+b+c=0$ hoặc $a=b=c$ $\Leftrightarrow a=b=c$(vì a,b,c là các số dương) (đpcm)
Pt đã cho $ (a+b)^{3}+c^{3}-3ab(a+b)-3abc=0$$\Leftrightarrow \left ( a+b+c \right )\left ( a^{2}+b^{2}+c^{2} -ab-bc-ca\right )=0$$\Leftrightarrow a+b+c=0$ hoặc $a=b=c$ $\Leftrightarrow a=b=c$(vì a,b,c là các số dương) (đpcm)
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sửa đổi
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giúp mình với !!!!!!!!!!!!!!!!
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Pt đã cho $\Leftrightarrow \left ( a+b \right )^{3}+c^{3} -3ab \left ( a+b \right )-3abc=0 $$\Leftrightarrow \left (a+b+c \right ) \left ( a^{2}+b^{2}+c^{2}+2ab-ac-bc\right )-3ab\left ( a+b+c \right )=0$$\Leftrightarrow \left ( a+b+c \right )\left ( a^{2}+b^{2}+c^{2} -ab-bc-ca\right )=0$$\Leftrightarrow a+b+c=0$ hoặc $a=b=c$ $\Leftrightarrow a=b=c$(vì a,b,c là các số dương) (đpcm)
Pt đã cho $ (a+b)^{3}+c^{3}-3ab(a+b)-3abc=0$$\Leftrightarrow \left ( a+b+c \right )\left ( a^{2}+b^{2}+c^{2} -ab-bc-ca\right )=0$$\Leftrightarrow a+b+c=0$ hoặc $a=b=c$ $\Leftrightarrow a=b=c$(vì a,b,c là các số dương) (đpcm)
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sửa đổi
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giúp mình với !!!!!!!!!!!!!!!!
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Pt đã cho $\Leftrightarrow \left ( a+b \right )^{3}+c^{3} -3ab \left ( a+b \right )-3abc=0 $$\Leftrightarrow \left (a+b+c \right ) \left ( a^{2}+b^{2}+c^{2}+2ab-ac-bc\right )-3ab\left ( a+b+c \right )=0$$\Leftrightarrow \left ( a+b+c \right )\left ( a^{2}+b^{2}+c^{2} -ab-bc-ca\right )=0$$\Leftrightarrow a+b+c=0$ hoặc $a=b=c$ $\Leftrightarrow a=b=c$(vì a,b,c là các số dương) (đpcm)
Pt đã cho $\Leftrightarrow \left ( a+b \right )^{3}+c^{3} -3ab \left ( a+b \right )-3abc=0 $$\Leftrightarrow \left (a+b+c \right ) \left ( a^{2}+b^{2}+c^{2}+2ab-ac-bc\right )-3ab\left ( a+b+c \right )=0$$\Leftrightarrow \left ( a+b+c \right )\left ( a^{2}+b^{2}+c^{2} -ab-bc-ca\right )=0$$\Leftrightarrow a+b+c=0$ hoặc $a=b=c$ $\Leftrightarrow a=b=c$(vì a,b,c là các số dương) (đpcm)
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sửa đổi
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giúp mình với !!!!!!!!!!!!!!!!
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Pt đã cho$\Leftrightarrow \left ( a+b \right )^{3}+c^{3} -3ab\left ( a+b \right )-3abc=0 $$\Leftrightarrow \left (a+b+c \right )\left ( a^{2}+b^{2}+c^{2}+2ab-ac-bc\right )-3ab\left ( a+b+c \right )=0$$\Leftrightarrow \left ( a+b+c \right )\left ( a^{2}+b^{2}+c^{2} -ab-bc-ca\right )=0$$\Leftrightarrow a+b+c=0$ hoặc $a=b=c$ $\Leftrightarrow a=b=c$(vì a,b,c là các số dương) (đpcm)
Pt đã cho $\Leftrightarrow \left ( a+b \right )^{3}+c^{3} -3ab \left ( a+b \right )-3abc=0 $$\Leftrightarrow \left (a+b+c \right ) \left ( a^{2}+b^{2}+c^{2}+2ab-ac-bc\right )-3ab\left ( a+b+c \right )=0$$\Leftrightarrow \left ( a+b+c \right )\left ( a^{2}+b^{2}+c^{2} -ab-bc-ca\right )=0$$\Leftrightarrow a+b+c=0$ hoặc $a=b=c$ $\Leftrightarrow a=b=c$(vì a,b,c là các số dương) (đpcm)
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