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Lè lè cho cái lưỡi dài ra!!!!!!!$pt\Leftrightarrow (\sqrt[3]{x}-\sqrt[3]{x-1})^3+(\sqrt[3]{x}-\sqrt[3]{x-1})=(x+1)+\sqrt[3]{x+1}$$\Leftrightarrow \sqrt[3]{x}-\sqrt[3]{x-1}=\sqrt[3]{x+1}$$\Leftrightarrow \sqrt[3] x=\sqrt[3]{x+1}+\sqrt[3]{x-1}$$\Leftrightarrow x=x+1+x-1+3(\sqrt[3]{x+1}+\sqrt[3]{x-1}).\sqrt[3]{(x+1)(x-1})$$\Leftrightarrow \frac{x+1+x-1}{2}+3(\sqrt[3]{x+1}+\sqrt[3]{x-1}).\sqrt[3]{(x+1)(x-1})=0$Đặt $\sqrt[3]{x+1}=a,\sqrt[3]{x-1}=b$ (a>b)$\Rightarrow \dfrac{a^3+b^3}2+3ab(a+b)=0 $$\Leftrightarrow (a+b)(a^2-ab+b^2+6ab)=0$$\Leftrightarrow \left[ \begin{array}{l} a+b=0\\ a^2+5ab+b^2=0 \end{array} \right.$Với $a+b=0$ dễ dàng tìm đc $x=0$Với $a^2+5ab+b^2=0$$\Leftrightarrow \left[ \begin{array}{l} a=\frac{b(\sqrt{21}-5)}{2} (1)\\ a=\frac{b(-\sqrt{21}-5)}{2} (2) \end{array} \right.$với (1) ta có : $2\sqrt[3]{x+1}=\sqrt[3]{x-1}(\sqrt{21}-5)$$\Leftrightarrow x=\frac{-8-(\sqrt{21}-5)^3}{8-(\sqrt{21}-5)^3}=\frac{-48(2\sqrt{21}-9)}{-32(3\sqrt{21}-14)}=-\sqrt{\frac{27}{28}}$ với (2) ta có $2\sqrt[3]{x+1}=\sqrt[3]{x-1}(-\sqrt{21}-5)$$\Leftrightarrow x=\frac{-8-(-\sqrt{21}-5)^3}{8-(-\sqrt{21}-5)^3}=\sqrt{\frac{27}{28}}$ Vậy $\color{red}{S=\{0,\pm \sqrt{\frac{27}{28}}\}} $

Lè lè cho cái lưỡi dài ra!!!!!!!$pt\Leftrightarrow (\sqrt[3]{x}-\sqrt[3]{x-1})^3+(\sqrt[3]{x}-\sqrt[3]{x-1})=(x+1)+\sqrt[3]{x+1}$$\Leftrightarrow \sqrt[3]{x}-\sqrt[3]{x-1}=\sqrt[3]{x+1}$$\Leftrightarrow \sqrt[3] x=\sqrt[3]{x+1}+\sqrt[3]{x-1}$$\Leftrightarrow x=x+1+x-1+3(\sqrt[3]{x+1}+\sqrt[3]{x-1}).\sqrt[3]{(x+1)(x-1})$$\Leftrightarrow \frac{x+1+x-1}{2}+3(\sqrt[3]{x+1}+\sqrt[3]{x-1}).\sqrt[3]{(x+1)(x-1})=0$Đặt $\sqrt[3]{x+1}=a,\sqrt[3]{x-1}=b$ (a>b)$\Rightarrow \dfrac{a^3+b^3}2+3ab(a+b)=0 $$\Leftrightarrow (a+b)(a^2-ab+b^2+6ab)=0$$\Leftrightarrow \left[ \begin{array}{l} a+b=0\\ a^2+5ab+b^2=0 \end{array} \right.$Với $a+b=0$ dễ dàng tìm đc $x=0$Với $a^2+5ab+b^2=0$$\Leftrightarrow \left[ \begin{array}{l} a=\frac{b(\sqrt{21}-5)}{2} (1)\\ a=\frac{b(-\sqrt{21}-5)}{2} (2) \end{array} \right.$với (1) ta có : $2\sqrt[3]{x+1}=\sqrt[3]{x-1}(\sqrt{21}-5)$$\Leftrightarrow x=\frac{-8-(\sqrt{21}-5)^3}{8-(\sqrt{21}-5)^3}=\frac{-48(2\sqrt{21}-9)}{-32(3\sqrt{21}-14)}=-\sqrt{\frac{27}{28}}$ với (2) ta có $2\sqrt[3]{x+1}=\sqrt[3]{x-1}(-\sqrt{21}-5)$$\Leftrightarrow x=\frac{-8-(-\sqrt{21}-5)^3}{8-(-\sqrt{21}-5)^3}=\sqrt{\frac{27}{28}}$ Vậy $\color{red}{S=\{0,\pm \sqrt{\frac{27}{28}}\}} $

Lè lè cho cái lưỡi dài ra!!!!!!!$pt\Leftrightarrow (\sqrt[3]{x}-\sqrt[3]{x-1})^3+(\sqrt[3]{x}-\sqrt[3]{x-1})=(x+1)+\sqrt[3]{x+1}$$\Leftrightarrow \sqrt[3]{x}-\sqrt[3]{x-1}=\sqrt[3]{x+1}$$\Leftrightarrow \sqrt[3] x=\sqrt[3]{x+1}+\sqrt[3]{x-1}$$\Leftrightarrow x=x+1+x-1+3(\sqrt[3]{x+1}+\sqrt[3]{x-1}).\sqrt[3]{(x+1)(x-1})$$\Leftrightarrow \frac{x+1+x-1}{2}+3(\sqrt[3]{x+1}+\sqrt[3]{x-1}).\sqrt[3]{(x+1)(x-1})=0$Đặt $\sqrt[3]{x+1}=a,\sqrt[3]{x-1}=b$ (a>b)$\Rightarrow \dfrac{a^3+b^3}2+3ab(a+b)=0 $$\Leftrightarrow (a+b)(a^2-ab+b^2+6ab)=0$$\Leftrightarrow \left[ \begin{array}{l} a+b=0\\ a^2+5ab+b^2=0 \end{array} \right.$Với $a+b=0$ dễ dàng tìm đc $x=0$Với $a^2+5ab+b^2$... chưa làm ra :)) (chỉ biết $x^2=\frac{27}{28} $ bạn nào làm thử)$a^2+5ab+b^2=0 $$<=> a=\frac{b(\sqrt{21}-5)}{2} (1)$hoặc $a=\frac{b(-5-\sqrt{21})}{2}(2)$với (1) ta có : $2\sqrt[3]{x+1}=\sqrt[3]{x-1}(\sqrt{21}-5)$$<=> x=\frac{-8-(\sqrt{21}-5)^3}{8-(\sqrt{21}-5)^3}$tương tự với (2)ps: đáp án $x^2=\frac{27}{28} $là đáp án gần đúng thôi

Lè lè cho cái lưỡi dài ra!!!!!!!$pt\Leftrightarrow (\sqrt[3]{x}-\sqrt[3]{x-1})^3+(\sqrt[3]{x}-\sqrt[3]{x-1})=(x+1)+\sqrt[3]{x+1}$$\Leftrightarrow \sqrt[3]{x}-\sqrt[3]{x-1}=\sqrt[3]{x+1}$$\Leftrightarrow \sqrt[3] x=\sqrt[3]{x+1}+\sqrt[3]{x-1}$$\Leftrightarrow x=x+1+x-1+3(\sqrt[3]{x+1}+\sqrt[3]{x-1}).\sqrt[3]{(x+1)(x-1})$$\Leftrightarrow \frac{x+1+x-1}{2}+3(\sqrt[3]{x+1}+\sqrt[3]{x-1}).\sqrt[3]{(x+1)(x-1})=0$Đặt $\sqrt[3]{x+1}=a,\sqrt[3]{x-1}=b$ (a>b)$\Rightarrow \dfrac{a^3+b^3}2+3ab(a+b)=0 $$\Leftrightarrow (a+b)(a^2-ab+b^2+6ab)=0$$\Leftrightarrow \left[ \begin{array}{l} a+b=0\\ a^2+5ab+b^2=0 \end{array} \right.$Với $a+b=0$ dễ dàng tìm đc $x=0$Với $a^2+5ab+b^2=0$$\Leftrightarrow \left[ \begin{array}{l} a=\frac{b(\sqrt{21}-5)}{2} (1)\\ a=\frac{b(-\sqrt{21}-5)}{2} (2) \end{array} \right.$với (1) ta có : $2\sqrt[3]{x+1}=\sqrt[3]{x-1}(\sqrt{21}-5)$$\Leftrightarrow x=\frac{-8-(\sqrt{21}-5)^3}{8-(\sqrt{21}-5)^3}=\frac{-48(2\sqrt{21}-9)}{-32(3\sqrt{21}-14)}=-\sqrt{\frac{27}{28}}$ với (2) ta có $2\sqrt[3]{x+1}=\sqrt[3]{x-1}(-\sqrt{21}-5)$$\Leftrightarrow x=\frac{-8-(-\sqrt{21}-5)^3}{8-(-\sqrt{21}-5)^3}=\sqrt{\frac{27}{28}}$ Vậy $\color{red}{S=\{0,\pm \sqrt{\frac{27}{28}}\}} $

Lè lè cho cái lưỡi dài ra!!!!!!!$pt\Leftrightarrow (\sqrt[3]{x}-\sqrt[3]{x-1})^3+(\sqrt[3]{x}-\sqrt[3]{x-1})=(x+1)+\sqrt[3]{x+1}$$\Leftrightarrow \sqrt[3]{x}-\sqrt[3]{x-1}=\sqrt[3]{x+1}$$\Leftrightarrow \sqrt[3] x=\sqrt[3]{x+1}+\sqrt[3]{x-1}$$\Leftrightarrow x=x+1+x-1+3(\sqrt[3]{x+1}+\sqrt[3]{x-1}).\sqrt[3]{(x+1)(x-1})$$\Leftrightarrow \frac{x+1+x-1}{2}+3(\sqrt[3]{x+1}+\sqrt[3]{x-1}).\sqrt[3]{(x+1)(x-1})=0$Đặt $\sqrt[3]{x+1}=a,\sqrt[3]{x-1}=b$ (a>b)$\Rightarrow \dfrac{a^3+b^3}2+3ab(a+b)=0 $$\Leftrightarrow (a+b)(a^2-ab+b^2+6ab)=0$$\Leftrightarrow \left[ \begin{array}{l} a+b=0\\ a^2+5ab+b^2=0 \end{array} \right.$Với $a+b=0$ dễ dàng tìm đc $x=0$Với $a^2+5ab+b^2$... chưa làm ra :)) (chỉ biết $x^2=\frac{27}{28} $ bạn nào làm thử)

Lè lè cho cái lưỡi dài ra!!!!!!!$pt\Leftrightarrow (\sqrt[3]{x}-\sqrt[3]{x-1})^3+(\sqrt[3]{x}-\sqrt[3]{x-1})=(x+1)+\sqrt[3]{x+1}$$\Leftrightarrow \sqrt[3]{x}-\sqrt[3]{x-1}=\sqrt[3]{x+1}$$\Leftrightarrow \sqrt[3] x=\sqrt[3]{x+1}+\sqrt[3]{x-1}$$\Leftrightarrow x=x+1+x-1+3(\sqrt[3]{x+1}+\sqrt[3]{x-1}).\sqrt[3]{(x+1)(x-1})$$\Leftrightarrow \frac{x+1+x-1}{2}+3(\sqrt[3]{x+1}+\sqrt[3]{x-1}).\sqrt[3]{(x+1)(x-1})=0$Đặt $\sqrt[3]{x+1}=a,\sqrt[3]{x-1}=b$ (a>b)$\Rightarrow \dfrac{a^3+b^3}2+3ab(a+b)=0 $$\Leftrightarrow (a+b)(a^2-ab+b^2+6ab)=0$$\Leftrightarrow \left[ \begin{array}{l} a+b=0\\ a^2+5ab+b^2=0 \end{array} \right.$Với $a+b=0$ dễ dàng tìm đc $x=0$Với $a^2+5ab+b^2$... chưa làm ra :)) (chỉ biết $x^2=\frac{27}{28} $ bạn nào làm thử)$a^2+5ab+b^2=0 $$<=> a=\frac{b(\sqrt{21}-5)}{2} (1)$hoặc $a=\frac{b(-5-\sqrt{21})}{2}(2)$với (1) ta có : $2\sqrt[3]{x+1}=\sqrt[3]{x-1}(\sqrt{21}-5)$$<=> x=\frac{-8-(\sqrt{21}-5)^3}{8-(\sqrt{21}-5)^3}$tương tự với (2)ps: đáp án $x^2=\frac{27}{28} $là đáp án gần đúng thôi

$pt\Leftrightarrow (\sqrt[3]{x}-\sqrt[3]{x-1})^3+(\sqrt[3]{x}-\sqrt[3]{x-1})=(x+1)+\sqrt[3]{x+1}$$\Leftrightarrow \sqrt[3]{x}-\sqrt[3]{x-1}=\sqrt[3]{x+1}$$\Leftrightarrow \sqrt[3] x=\sqrt[3]{x+1}+\sqrt[3]{x-1}$$\Leftrightarrow x=x+1+x-1+3(\sqrt[3]{x+1}+\sqrt[3]{x-1}).\sqrt[3]{(x+1)(x-1})$$\Leftrightarrow \frac{x+1+x-1}{2}+3(\sqrt[3]{x+1}+\sqrt[3]{x-1}).\sqrt[3]{(x+1)(x-1})=0$Đặt $\sqrt[3]{x+1}=a,\sqrt[3]{x-1}=b$ (a>b)$\Rightarrow \dfrac{a^3+b^3}2+3ab(a+b)=0 $$\Leftrightarrow (a+b)(a^2-ab+b^2+6ab)=0$$\Leftrightarrow \left[ \begin{array}{l} a+b=0\\ a^2+5ab+b^2=0 \end{array} \right.$Với $a+b=0$ dễ dàng tìm đc $x=0$Với $a^2+5ab+b^2$... chưa làm ra :)) (chỉ biết $x^2=\frac{27}{28} $ bạn nào làm thử)

Lè lè cho cái lưỡi dài ra!!!!!!!$pt\Leftrightarrow (\sqrt[3]{x}-\sqrt[3]{x-1})^3+(\sqrt[3]{x}-\sqrt[3]{x-1})=(x+1)+\sqrt[3]{x+1}$$\Leftrightarrow \sqrt[3]{x}-\sqrt[3]{x-1}=\sqrt[3]{x+1}$$\Leftrightarrow \sqrt[3] x=\sqrt[3]{x+1}+\sqrt[3]{x-1}$$\Leftrightarrow x=x+1+x-1+3(\sqrt[3]{x+1}+\sqrt[3]{x-1}).\sqrt[3]{(x+1)(x-1})$$\Leftrightarrow \frac{x+1+x-1}{2}+3(\sqrt[3]{x+1}+\sqrt[3]{x-1}).\sqrt[3]{(x+1)(x-1})=0$Đặt $\sqrt[3]{x+1}=a,\sqrt[3]{x-1}=b$ (a>b)$\Rightarrow \dfrac{a^3+b^3}2+3ab(a+b)=0 $$\Leftrightarrow (a+b)(a^2-ab+b^2+6ab)=0$$\Leftrightarrow \left[ \begin{array}{l} a+b=0\\ a^2+5ab+b^2=0 \end{array} \right.$Với $a+b=0$ dễ dàng tìm đc $x=0$Với $a^2+5ab+b^2$... chưa làm ra :)) (chỉ biết $x^2=\frac{27}{28} $ bạn nào làm thử)