Giải câu 7 trang 33 toán VNEN 9 tập 1.
a) P = $\frac{1}{2\sqrt{x} - 2}$ - $\frac{1}{2\sqrt{x} + 2}$ + $\frac{\sqrt{x}}{1 - x}$
= $\frac{1}{2(\sqrt{x} - 1)}$ - $\frac{1}{2(\sqrt{x} + 1)}$ + $\frac{\sqrt{x}}{1 - x}$
= $\frac{\sqrt{x} + 1}{2(\sqrt{x} - 1)(\sqrt{x} + 1)}$ - $\frac{\sqrt{x} - 1}{2(\sqrt{x} + 1)(\sqrt{x} - 1)}$ + $\frac{\sqrt{x}}{1 - x}$
= $\frac{\sqrt{x} + 1 - \sqrt{x} + 1}{2(\sqrt{x} - 1)(\sqrt{x} + 1)}$ + $\frac{\sqrt{x}}{1 - x}$
= $\frac{2}{2(x - 1)}$ + $\frac{\sqrt{x}}{1 - x}$
= $\frac{1}{x - 1}$ - $\frac{\sqrt{x}}{x - 1}$
= $\frac{1 - \sqrt{x}}{(\sqrt{x} - 1)(\sqrt{x} + 1)}$
= - $\frac{1}{\sqrt{x} + 1}$
b) Với x = $\frac{4}{9}$ thì P = - $\frac{3}{5}$
c) $\left | P \right |$ = $\frac{1}{3}$ $\Leftrightarrow $ $\left | \frac{- 1}{\sqrt{x} + 1} \right |$ = $\frac{1}{3}$
Vì $\sqrt{x}$ + 1 > 0 nên $\frac{1}{\sqrt{x} + 1}$ > 0
Khi đó $\left | \frac{- 1}{\sqrt{x} + 1} \right |$ = $\frac{1}{3}$ $\Leftrightarrow $ $\frac{1}{\sqrt{x} + 1}$ = $\frac{1}{3}$ $\Leftrightarrow $ x = 4
Vậy x = 4.