Giải câu 1 đề 4 ôn thi toán lớp 9 lên 10

$A=\left ( \frac{x-5\sqrt{x}}{x - 25}-1 \right ):\left ( \frac{25-x}{x+2\sqrt{x}-15}--\frac{\sqrt{x}+3}{\sqrt{x}+5}+\frac{\sqrt{x}-5}{\sqrt{x}-3} \right )$

$=\left ( \frac{\sqrt{x} (\sqrt{x}-5)}{(\sqrt{x}-5)(\sqrt{x}+5)-1}\right ):\left [ \frac{25-x-(\sqrt{x}+3)(\sqrt{x}-3)+(\sqrt{x}-5)(\sqrt{x}+5)}{(\sqrt{x}+5)(\sqrt{x}-5)} \right ]$

$=\frac{-5}{\sqrt{x}+5}:\frac{9-x}{(\sqrt{5}+x)(\sqrt{x}-3)}$

$=\frac{-5}{\sqrt{x}+5}.\frac{(\sqrt{x}+5)(\sqrt{x}-3)}{(\sqrt{x}+3)(3-\sqrt{x})}$

$=\frac{5}{\sqrt{x}+3}$

b. $A <1\Leftrightarrow \frac{5}{\sqrt{X}+3}< 1\Leftrightarrow \frac{5-(\sqrt{X}+3)}{\sqrt{X}+3}< 0$

$\Leftrightarrow \frac{2-\sqrt{x}}{\sqrt{x}+3}< 0 \Leftrightarrow 2-\sqrt{x} < 0 (Do \sqrt{x} + 3 > 0)$

$\Leftrightarrow \sqrt{x} >2 \Leftrightarrow x > 4$

Vậy với x > 4, x # 9, x # 25 thì A < 1