Đáp án câu 1 đề 9 kiểm tra học kì 2 Toán 9

a, Khi x = $\frac{4}{9}$ (thỏa mãn đkxđ) thì $\sqrt{x} = \frac{2}{3}$. Khi đó:

A = $\frac{\frac{2}{3}}{ 1 + 3.\frac{2}{3}} = \frac{2}{9}$

b, Với $x \geq 0; x \neq 9$, ta có:

B = $\frac{x + 3}{x - 9} + \frac{2}{\sqrt{x} + 3} - \frac{1}{3 - \sqrt{x}}$

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

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

  = $\frac{x + 3\sqrt{x}}{(\sqrt{x} - 3)(\sqrt{x} + 3)}$

  = $\frac{\sqrt{x}(\sqrt{x} + 3)}{(\sqrt{x} - 3)(\sqrt{x} + 3)}$

  = $\frac{\sqrt{x}}{\sqrt{x} - 3}$

c, Ta có:

P = B : A = $\frac{\sqrt{x}}{\sqrt{x} - 3} : \frac{\sqrt{x}}{1 + 3\sqrt{x}}= \frac{1 + 3\sqrt{x}}{\sqrt{x} - 3}$

$P < 3 \Rightarrow  \frac{1 + 3\sqrt{x}}{\sqrt{x} - 3} < 3$

$\Leftrightarrow \frac{3(\sqrt{x} - 3) + 10}{\sqrt{x} - 3} < 3$

$\Leftrightarrow 3 + \frac{10}{\sqrt{x} - 3} < 3$

$\Leftrightarrow \frac{10}{\sqrt{x} - 3} < 0$

$\Leftrightarrow \sqrt{x} - 3 < 0$

$\Leftrightarrow \sqrt{x} < 3$

$\Leftrightarrow x < 9$

Mà $x \geq 0; x \neq 9$ 

Vậy 0 < x < 9.