$Nhớ....................Vote....................Nha$ Dk: $x\ge -1=>x+1\ge 0$$pt\iff 5(\sqrt{1+x^3}-2x-2)=4x^4-25x^3+18x^2-10x-15$$\iff \frac{5(1+x^3-4x^2-8x-4)}{\sqrt{1+x^3}+2x+2}=(x^2-5x-3)(4x^2-5x+5)$$\iff \frac{5(x+1)(x^2-5x-3)}{\sqrt{1+x^3}+2x+2}=(x^2-5x-3)(4x^2-5x+5)$$x^2-5x-3=0.v.\frac{5(x+1)}{\sqrt{1+x^3}+2x+2}=4x^2-5x+5 $$Th1: x^2-5x-3=0=>.....$Ta có: $4x^2-5x+5=(2x-\frac{5}{4})^2+\frac{55}{16}>3$$Th2:\frac{5(x+1)}{\sqrt{1+x^3}+2x+2}=4x^2-5x+5>3=>5(x+1)>3\sqrt{1+x^3}+6x+6$$\iff 0>3\sqrt{1+x^3}+x+1(Vo.li)=>Vo.nghiem$Vậy $x=\frac{5+\sqrt{37}}{2},x=\frac{5-\sqrt{37}}{2}$
$Nhớ....................Vote....................Nha$Dk: $x\ge -1=>x+1\ge 0$$pt\iff 5(\sqrt{1+x^3}-2x-2)=4x^4-25x^3+18x^2-10x-15$$\iff \frac{5(1+x^3-4x^2-8x-4)}{\sqrt{1+x^3}+2x+2}=(x^2-5x-3)(4x^2-5x+5)$$\iff \frac{5(x+1)(x^2-5x-3)}{\sqrt{1+x^3}+2x+2}=(x^2-5x-3)(4x^2-5x+5)$$x^2-5x-3=0.v.\frac{5(x+1)}{\sqrt{1+x^3}+2x+2}=4x^2-5x+5 $$Th1: x^2-5x-3=0=>.....$Ta có: $4x^2-5x+5=(2x-\frac{5}{4})^2+\frac{55}{16}>3$$Th2:\frac{5(x+1)}{\sqrt{1+x^3}+2x+2}=4x^2-5x+5>3=>5(x+1)>3\sqrt{1+x^3}+6x+6$$\iff 0>3\sqrt{1+x^3}+x+1(Vo.li)=>Vo.nghiem$Vậy $x=\frac{5+\sqrt{37}}{2},x=\frac{5-\sqrt{37}}{2}$
$Nhớ....................Vote....................Nha$
Dk: $x\ge -1=>x+1\ge 0$$pt\iff 5(\sqrt{1+x^3}-2x-2)=4x^4-25x^3+18x^2-10x-15$$\iff \frac{5(1+x^3-4x^2-8x-4)}{\sqrt{1+x^3}+2x+2}=(x^2-5x-3)(4x^2-5x+5)$$\iff \frac{5(x+1)(x^2-5x-3)}{\sqrt{1+x^3}+2x+2}=(x^2-5x-3)(4x^2-5x+5)$$x^2-5x-3=0.v.\frac{5(x+1)}{\sqrt{1+x^3}+2x+2}=4x^2-5x+5 $$Th1: x^2-5x-3=0=>.....$Ta có: $4x^2-5x+5=(2x-\frac{5}{4})^2+\frac{55}{16}>3$$Th2:\frac{5(x+1)}{\sqrt{1+x^3}+2x+2}=4x^2-5x+5>3=>5(x+1)>3\sqrt{1+x^3}+6x+6$$\iff 0>3\sqrt{1+x^3}+x+1(Vo.li)=>Vo.nghiem$Vậy $x=\frac{5+\sqrt{37}}{2},x=\frac{5-\sqrt{37}}{2}$