Question

Answer 1

Answer 2

a: Ta có: \(P=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\)

\(=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

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

\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)

b: Ta có: \(x^2=\dfrac{16}{81}\)

nên \(\left[{}\begin{matrix}x=\dfrac{4}{9}\left(nhận\right)\\x=-\dfrac{4}{9}\left(loại\right)\end{matrix}\right.\)

Thay \(x=\dfrac{4}{9}\) vào P, ta được:

\(P=\left(\dfrac{2}{3}-1\right):\left(\dfrac{2}{3}+1\right)=\dfrac{-1}{3}:\dfrac{5}{3}=-\dfrac{1}{5}\)