Đáp án:Giải thích các bước giải:
a) $P=(\dfrac{\sqrt[]{x}-1}{3\sqrt[]{x}-1}-$ $\dfrac{1}{3\sqrt[]{x}+1}+$ $\dfrac{5\sqrt[]{x}}{9x-1}):(1-$ $\dfrac{3\sqrt[]{x}-2}{3\sqrt[]{x}+1})$
$P=(\dfrac{(\sqrt[]{x}-1).3\sqrt[]{x}+1-3\sqrt[]{x}+1+8\sqrt[]{x}}{(3\sqrt[]{x}-1)(3\sqrt[]{x}+1)}):($ $\dfrac{3\sqrt[]{x}+1-3\sqrt[]{x}+2}{3\sqrt[]{x}+1})$
$P=(\dfrac{3x+\sqrt[]{x}-3\sqrt[]{x}-1+5\sqrt[]{x}+1}{(3\sqrt[]{x}-1)(3\sqrt[]{x}+1)}:$ $\dfrac{3}{3\sqrt[]{x}+1}$
$P=\dfrac{3x+3\sqrt[]{x}}{(3\sqrt[]{x}-1)(3\sqrt[]{x}+1)}.$ $\dfrac{3\sqrt[]{x}+1}{3}$
$P=\dfrac{3\sqrt[]{x}.(\sqrt[]{x}+1).(3\sqrt[]{x}+1)}{3.(3\sqrt[]{x}-1)(3\sqrt[]{x}+1)}$
$P=\dfrac{x+\sqrt[]{x}}{3\sqrt[]{x}-1}$
b) để P= $\dfrac{6}{5}$
$⇒\dfrac{x+\sqrt[]{x}}{3\sqrt[]{x}-1}=$$\dfrac{6}{5}$
$\text{⇔ 5 ( x + }$$\sqrt[]{x})=6(3$$\sqrt[]{x}-1)$
$\text{⇔ 5x + 5}$$\sqrt[]{x}=$ $18\sqrt[]{x}-6$
$\text{⇔ 5x + 13}$$\sqrt[]{x}+6=0$
\(⇔\left[ \begin{array}{l}\sqrt[]{x}=2\\\sqrt[]{x}=\dfrac{3}{5}\end{array} \right.\) \(⇔\left[ \begin{array}{l}x=4(tm)\\x=\dfrac{9}{25}(tm)\end{array} \right.\)