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$\dfrac{1}{\left(x-1\right)^2}+\sqrt{3x+1}=\dfrac{1}{x^2}+\sqrt{x+2}$

Nguyễn Việt Lâm · Accepted Answer

ĐKXĐ: $x\ge\dfrac{-1}{3};x\ne0;1$

$\dfrac{1}{\left(x-1\right)^2}-\dfrac{1}{x^2}+\sqrt{3x+1}-\sqrt{x+2}=0$

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

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

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

$\Leftrightarrow2x-1=0$ (do $\dfrac{1}{\left(x^2-x\right)^2}+\dfrac{1}{\sqrt{3x+1}+\sqrt{x+2}}>0$ )

$\Rightarrow x=\dfrac{1}{2}$Câu trả lời: ĐKXĐ: $x\ge\dfrac{-1}{3};x\ne0;1$

$\dfrac{1}{\left(x-1\right)^2}-\dfrac{1}{x^2}+\sqrt{3x+1}-\sqrt{x+2}=0$

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

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

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

$\Leftrightarrow2x-1=0$ (do $\dfrac{1}{\left(x^2-x\right)^2}+\dfrac{1}{\sqrt{3x+1}+\sqrt{x+2}}>0$ )

$\Rightarrow x=\dfrac{1}{2}$

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