Câu hỏi của hakito

Question

Giải PT này giùm mk:

$\sqrt[3]{x^2-1}+x=\sqrt{x^3-2}$

Nguyen · Accepted Answer

ĐK: $x\ge\sqrt[3]{2}$

$\Leftrightarrow\sqrt{x^3-2}-\sqrt[3]{x^2-1}-x=0$

$\Leftrightarrow\sqrt{x^3-2}-5-\sqrt[3]{x^2-1}+2-x+3=0$

$\Leftrightarrow\dfrac{x^3-27}{\sqrt{x^3-2}+5}-\dfrac{x^2-1-8}{\sqrt[3]{x^2-1}^2+2\sqrt[3]{x^2-1}+4}-\left(x-3\right)=0$

$\Leftrightarrow\left(x-3\right)\left[\dfrac{x^2+3x+9}{\sqrt{x^3-2}+5}-\dfrac{x+3}{\sqrt[3]{x^2-1}^2+2\sqrt[3]{x^2-1}+4}-1\right]=0$

$\Leftrightarrow\left[{}\begin{matrix}x=3\\dfrac{x^2+3x+9}{\sqrt{x^3-2}+5}-\dfrac{x+3}{\sqrt[3]{x^2-1}^2+2\sqrt[3]{x^2-1}+4}-1=0\end{matrix}\right.$

Vì $\dfrac{x^2+3x+9}{\sqrt{x^3-2}+5}-\dfrac{x+3}{\sqrt[3]{x^2-1}^2+2\sqrt[3]{x^2-1}+4}-1>0\forall x\ge\sqrt[3]{2}$

nên x=3.Câu trả lời: ĐK: $x\ge\sqrt[3]{2}$

$\Leftrightarrow\sqrt{x^3-2}-\sqrt[3]{x^2-1}-x=0$

$\Leftrightarrow\sqrt{x^3-2}-5-\sqrt[3]{x^2-1}+2-x+3=0$

$\Leftrightarrow\dfrac{x^3-27}{\sqrt{x^3-2}+5}-\dfrac{x^2-1-8}{\sqrt[3]{x^2-1}^2+2\sqrt[3]{x^2-1}+4}-\left(x-3\right)=0$

$\Leftrightarrow\left(x-3\right)\left[\dfrac{x^2+3x+9}{\sqrt{x^3-2}+5}-\dfrac{x+3}{\sqrt[3]{x^2-1}^2+2\sqrt[3]{x^2-1}+4}-1\right]=0$

$\Leftrightarrow\left[{}\begin{matrix}x=3\\dfrac{x^2+3x+9}{\sqrt{x^3-2}+5}-\dfrac{x+3}{\sqrt[3]{x^2-1}^2+2\sqrt[3]{x^2-1}+4}-1=0\end{matrix}\right.$

Vì $\dfrac{x^2+3x+9}{\sqrt{x^3-2}+5}-\dfrac{x+3}{\sqrt[3]{x^2-1}^2+2\sqrt[3]{x^2-1}+4}-1>0\forall x\ge\sqrt[3]{2}$

nên x=3.

hakito · Answer

@Akai Haruma giải giúp em ạCâu trả lời: @Akai Haruma giải giúp em ạ

Violympic toán 9