Đáp án:
`a.`Tìm `x`:
Để `A=1` thì `\frac{2}{3-sqrtx}=1` (`x>0;x\ne4`)
`<=> 2=3-sqrtx`
`<=> sqrtx=3-2`
`<=> sqrtx=1`
`<=> x=1` (tm)
`b.`Tìm `x`:
Để `M=sqrtx` thì `\frac{sqrtx+1}{sqrtx-1}=sqrtx` (`x>1;x\ne1;x>1`)
`<=> sqrtx+1=(sqrtx-1)sqrtx`
`<=> sqrtx+1=x-sqrtx`
`<=> sqrt+sqrtx=x-1`
`<=> 2sqrtx=x-1`
`<=> 4x-x^2+2x-1=0`
`<=> 6x-x^2-1=0`
`<=> -x^2+6x-1=0`
`<=> x^2-6x+1=0`
`<=> x^2-6x=-1`
`<=> x^2-6x+9=-1+9`
`<=> (x-3)^2=8`
`<=>`\(\left[ \begin{array}{l}x=-2\sqrt{2}+3\ (ktm) \\x=2\sqrt{2}+3 \ (tm)\end{array} \right.\)
`<=> x=2sqrt2+3`
`c.`Tìm `a`:
Để `Q=-2` thì `\frac{1-a}{a}=-2` (`a\ne0`)
`<=> 1-a=-2a`
`<=> 1-a+2a=0`
`<=> 1+a=0`
`<=> a=-1` (tm)