P= ($\frac{1}{\sqrt{x}-\sqrt{x-1}}$ – $\frac{x-3}{\sqrt{x-1}-\sqrt{2}}$ )($\frac{2}{\sqrt{2}-\sqrt{x}}$- $\frac{\sqrt{x}+\sqrt{2}}{\sqrt{2x} -x}$)

0 bình luận về “P= ($\frac{1}{\sqrt{x}-\sqrt{x-1}}$ – $\frac{x-3}{\sqrt{x-1}-\sqrt{2}}$ )($\frac{2}{\sqrt{2}-\sqrt{x}}$- $\frac{\sqrt{x}+\sqrt{2}}{\sqrt{2x} -x}$)”

1. Đáp án: $P=\dfrac{\sqrt{2}-\sqrt{x}}{\sqrt{x}}$

   Giải thích các bước giải:

   Ta có:

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

   $\to P=\left(\dfrac{x-\left(x-1\right)}{\sqrt{x}-\sqrt{x-1}}-\dfrac{\left(x-1\right)-2}{\sqrt{x-1}-\sqrt{2}}\right)\cdot\left(\dfrac{2}{\sqrt{2}-\sqrt{x}}-\dfrac{\sqrt{x}+\sqrt{2}}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}\right)$

   $\to P=\left(\dfrac{\left(\sqrt{x}-\sqrt{x-1}\right)\cdot\left(\sqrt{x}+\sqrt{x-1}\right)}{\sqrt{x}-\sqrt{x-1}}-\dfrac{\left(\sqrt{x-1}-\sqrt{2}\right)\cdot\left(\sqrt{x-1}+\sqrt{2}\right)}{\sqrt{x-1}-\sqrt{2}}\right)\cdot\left(\dfrac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}-\dfrac{\sqrt{x}+\sqrt{2}}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}\right)$

   $\to P=\left(\left(\sqrt{x}+\sqrt{x-1}\right)-\left(\sqrt{x-1}+\sqrt{2}\right)\right)\cdot\left(\dfrac{2\sqrt{x}-\sqrt{x}-\sqrt{2}}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}\right)$

   $\to P=\left(\sqrt{x}-\sqrt{2}\right)\cdot\left(\dfrac{\sqrt{x}-\sqrt{2}}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}\right)$

   $\to P=\dfrac{\sqrt{2}-\sqrt{x}}{\sqrt{x}}$

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