Toán Lớp 8: x³ + $\sqrt{(x+1)^3}$ = 9x + 8
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Toán Lớp 8: x³ + $\sqrt{(x+1)^3}$ = 9x + 8
Comments ( 1 )
$ĐK:x\ge -1$
$PT\Leftrightarrow x^3+1+\sqrt{(x+1)^3}-9(x+1)=0$
$\Leftrightarrow (x+1)(x^2-x+1)+(x+1)\sqrt{x+1}-9(x+1)=0$
$\Leftrightarrow (x+1)(x^2-x+1+\sqrt{x+1}-9)=0$
$\Leftrightarrow (x+1)(x^2-x-8+\sqrt{x+1})=0$
$\Leftrightarrow (x+1)[x(x-3)+(\sqrt{x+1}-2)+2(x-3)]=0$
$\Leftrightarrow (x+1)[(x+2)(x-3)+\dfrac{x-3}{\sqrt{x+1}+2}]=0$
$\Leftrightarrow (x+1)(x-3)(x+2+\dfrac{1}{\sqrt{x+1}+2})=0$
Với $x\ge -1$ ta luôn có $x+2>0\Rightarrow x+2+\dfrac{1}{\sqrt{x+1}+2}>0$
Do đó PT tương đương $(x+1)(x-3)=0$
$\Leftrightarrow$\(\left[ \begin{array}{l}x=-1(tm)\\x=3(tm)\end{array} \right.\)
Vậy $S=\text{{-1;3}}$