Đáp án:
`A=\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}` với $a;b>0;a\neq b$
$B=\sqrt{x}-1$ với $x≥0;x\neq 1$
$C=4x$ với $x>0;x\neq 1$
Giải thích các bước giải:
$a)ĐKXĐ:a;b>0;a\neq b$
`A=(\frac{\sqrt{a}}{\sqrt{ab}-b}+\frac{2\sqrt{a}-\sqrt{b}}{\sqrt{ab}-a})÷(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}})`
`=[\frac{\sqrt{a}}{\sqrt{b}(\sqrt{a}-\sqrt{b})}+\frac{2\sqrt{a}-\sqrt{b}}{\sqrt{a}(\sqrt{b}-\sqrt{a})}]÷\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}.\sqrt{b}}`
`=\frac{\sqrt{a}.\sqrt{a}-(2\sqrt{a}-\sqrt{b})\sqrt{b}}{\sqrt{a}.\sqrt{b}(\sqrt{a}-\sqrt{b})}.\frac{\sqrt{a}.\sqrt{b}}{\sqrt{a}+\sqrt{b}}`
`=\frac{a-2\sqrt{a}.\sqrt{b}+b}{\sqrt{a}.\sqrt{b}(\sqrt{a}-\sqrt{b})}.\frac{\sqrt{a}.\sqrt{b}}{\sqrt{a}+\sqrt{b}}`
`=\frac{(\sqrt{a}-\sqrt{b})^2.\sqrt{a}.\sqrt{b}}{\sqrt{a}.\sqrt{b}(\sqrt{a}-\sqrt{b})}`
`=\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}`
$b)ĐKXĐ:x≥0;x\neq 1$
`B=(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{4\sqrt{x}}{x+\sqrt{x}+1}-\frac{2\sqrt{x}+1}{x\sqrt{x}-1})(\sqrt{x}+\frac{2\sqrt{x}+1}{\sqrt{x}-1})`
`=\frac{\sqrt{x}(x+\sqrt{x}+1)-4\sqrt{x}(\sqrt{x}-1)-(2\sqrt{x}+1)}{(\sqrt{x}-1)(x+\sqrt{x}+1)}.\frac{\sqrt{x}(\sqrt{x}-1)+(2\sqrt{x}+1)}{\sqrt{x}-1}`
`=\frac{x\sqrt{x}+x+\sqrt{x}-4x+4\sqrt{x}-2\sqrt{x}-1}{(\sqrt{x}-1)(x+\sqrt{x}+1)}.\frac{x-\sqrt{x}+2\sqrt{x}+1}{\sqrt{x}-1}`
`=\frac{x\sqrt{x}-3x+3\sqrt{x}-1}{(\sqrt{x}-1)(x+\sqrt{x}+1)}.\frac{x+\sqrt{x}+1}{\sqrt{x}-1}`
`=\frac{(\sqrt{x}-1)^3(x+\sqrt{x}+1)}{(\sqrt{x}-1)^2(x+\sqrt{x}+1)}`
`=\sqrt{x}-1`
$c)ĐKXĐ:x>0;x\neq 1$
`C=(\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}+4\sqrt{x})(\sqrt{x}-\frac{1}{\sqrt{x}})`
`=\frac{(\sqrt{x}+1)^2-(\sqrt{x}-1)^2+4\sqrt{x}(x-1)}{(\sqrt{x}-1)(\sqrt{x}+1)}.\frac{x-1}{\sqrt{x}}`
`=\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1+4x\sqrt{x}-4\sqrt{x}}{x-1}.\frac{x-1}{\sqrt{x}}`
`=\frac{4x\sqrt{x}(x-1)}{(x-1)\sqrt{x}}=4x`
