Question

\((\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1})\cdot\frac{(1-x)^2}{2}\)

a, Rút gọn A

b, Tìm x để A dương

Answer 1

\(A=\left(\frac{\sqrt{X}-2}{X-1}-\frac{\sqrt{X}+2}{X+2\sqrt{X}+1}\right).\frac{\left(1-X\right)^2}{2}\)

=\(\left(\frac{\sqrt{X}-2}{X-1}-\frac{\sqrt{X}+2}{\left(\sqrt{X}+1\right)^2}\right).\frac{\left(1-X\right)^2}{2}\)

=\(\frac{\left(\sqrt{X}-2\right)\left(\sqrt{X}+1\right)^2-\left(X-1\right)\left(\sqrt{X}+2\right)}{\left(X-1\right)\left(\sqrt{X}+1\right)^2}.\frac{\left(X-1\right)^2}{-2}\)

=\(\frac{X\sqrt{X}+2X+\sqrt{X}-2X-4\sqrt{X}-2-X\sqrt{X}-2X+\sqrt{X}+2}{\left(\sqrt{X}+1\right)^2}.\frac{X-1}{-2}\)

=\(\frac{-2\sqrt{X}-2X}{\left(\sqrt{X}+1\right)^2}.\frac{X-1}{-2}\)

=\(\frac{-2\sqrt{X}\left(1+\sqrt{X}\right)}{\left(\sqrt{X}+1\right)^2}.\frac{X-1}{-2}\)

=\(\frac{\sqrt{X}\left(1+\sqrt{X}\right)\left(\sqrt{X}+1\right)\left(\sqrt{X}-1\right)}{\left(\sqrt{X}+1\right)^2}\)

=\(\sqrt{X}\left(\sqrt{X}-1\right)\)

Answer 2

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b) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne\pm1\end{matrix}\right.\)

Để A dương \(\Leftrightarrow A>0\) \(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)>0\) \(\Leftrightarrow\left[{}\begin{matrix}x< 0\left(loại\right)\\x>1\left(TM\right)\end{matrix}\right.\)

Vậy \(x>1\) thì A dương