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Giải thích các bước giải:
Ta có:
$A=\left(\dfrac{x+2}{x-1}+\dfrac{\sqrt{x}}{1-\sqrt{x}}+\dfrac{1}{\sqrt{x}+1}\right).\left(\dfrac{x\sqrt{x}-1}{\sqrt{x}-1}+\sqrt{x}\right)$
$\to A=\left(\dfrac{x+2}{x-1}-\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right).\left(\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}-1}+\sqrt{x}\right)$
$\to A=\left(\dfrac{x+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right).\left(x+\sqrt{x}+1+\sqrt{x}\right)$
$\to A=\left(\dfrac{x+2-\sqrt{x}\left(\sqrt{x}+1\right)+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right).\left(x+2\sqrt{x}+1\right)$
$\to A=\left(\dfrac{x+2-\left(x+\sqrt{x}\right)+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right).\left(\sqrt{x}+1\right)^2$
$\to A=\left(\dfrac{1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right).\left(\sqrt{x}+1\right)^2$
$\to A=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}$
Hãy giúp mọi người biết câu trả lời này thế nào?
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