Đáp án: $P=1$
Giải thích các bước giải:
$\text{Do a<b}$
$⇒2a<2b$
$⇒0<2a-b<b$
$⇒\sqrt{2a-b}<\sqrt{b}$
$⇒\sqrt{2a-b}-\sqrt{b}<0$
$⇒\sqrt{(\sqrt{2a-b}-\sqrt{b})^2}=|\sqrt{2a-b}-\sqrt{b}|=\sqrt{b}-\sqrt{2a-b}$
$\text{Ta có:}$
`1+ax=1+a.\frac{1}{a}.\sqrt{\frac{2a-b}{b}}`
`=1+\frac{\sqrt{2a-b}}{\sqrt{b}}`
`=\frac{\sqrt{2a-b}+\sqrt{b}}{\sqrt{b}}`
`1-ax=1-a.\frac{1}{a}.\sqrt{\frac{2a-b}{b}}`
`=1-\frac{\sqrt{2a-b}}{\sqrt{b}}`
`=\frac{\sqrt{b}-\sqrt{2a-b}}{\sqrt{b}}`
`⇒\frac{1+ax}{1-ax}=\frac{\frac{\sqrt{2a-b}+\sqrt{b}}{\sqrt{b}}}{\frac{\sqrt{a}-\sqrt{2a-b}}{\sqrt{b}}}`
`=\frac{\sqrt{2a-b}+\sqrt{b}}{\sqrt{b}-\sqrt{2a-b}}`
$\text{Lại có:}$
`1-bx=1-b.\frac{1}{a}\sqrt{\frac{2a-b}{b}}`
`=1-\frac{b}{a}.\frac{\sqrt{2a-b}}{\sqrt{b}}`
`=1-\frac{\sqrt{b}.\sqrt{2a-b}}{a}`
`=\frac{a-\sqrt{b}.\sqrt{2a-b}}{a}`
`1+bx=1-b.\frac{1}{a}\sqrt{\frac{2a-b}{b}}`
`=1-\frac{b}{a}.\frac{\sqrt{2a-b}}{\sqrt{b}}`
`=1-\frac{\sqrt{b}.\sqrt{2a-b}}{a}`
`=\frac{a+\sqrt{b}.\sqrt{2a-b}}{a}`
`⇒\frac{1-bx}{1+bx}=\frac{\frac{a-\sqrt{b}.\sqrt{2a-b}}{a}}{\frac{a+\sqrt{b}.\sqrt{2a-b}}{a}}`
`=\frac{a-\sqrt{b}.\sqrt{2a-b}}{a+\sqrt{b}.\sqrt{2a-b}}`
`=\frac{2a-2\sqrt{b}.\sqrt{2a-b}}{2a+2\sqrt{b}.\sqrt{2a-b}}`
`=\frac{(2a-b)-2\sqrt{b}.\sqrt{2a-b}+b}{(2a-b)+2\sqrt{b}.\sqrt{2a-b}+b}`
`=\frac{(\sqrt{2a-b}-\sqrt{b})^2}{(\sqrt{b}+\sqrt{2a-b})^2}`
`⇒\sqrt{\frac{1-bx}{1+bx}}=\sqrt{\frac{(\sqrt{2a-b}-\sqrt{b})^2}{(\sqrt{b}+\sqrt{2a-b})^2}}`
`=\frac{\sqrt{b}-\sqrt{2a-b}}{\sqrt{b}+\sqrt{2a-b}}`
$\text{Ta có:}$
`P=\frac{1+ax}{1-ax}.\sqrt{\frac{1-bx}{1+bx}}`
`=\frac{\sqrt{2a-b}+\sqrt{b}}{\sqrt{b}-\sqrt{2a-b}}.\frac{\sqrt{b}-\sqrt{2a-b}}{\sqrt{b}+\sqrt{2a-b}}`
$=1$
