Đáp án + Giải thích các bước giải:
a.
`A=(\frac{sqrta-asqrta}{a+2sqrta+1}-sqrta):\frac{1-1/sqrta}{sqrta}`
`=[\frac{sqrta-asqrta-sqrta(a+2sqrta+1)}{a+2sqrta+1}]:\frac{(sqrta-1)/sqrta}{sqrta}`
`=\frac{-2asqrta-2a}{a+2sqrta+1}:\frac{sqrta-1}{a}`
`=\frac{-2a(sqrta+1)}{(sqrta+1)^2}.\frac{a}{sqrta-1}`
`=\frac{-2a}{sqrta+1}.\frac{a}{sqrta-1}`
`=\frac{-2a^2}{(sqrta+1)(sqrta-1)}`
`=\frac{-2a^2}{a-1}`
b.
Để `|A|=1`
`<=>|\frac{-2a^2}{a-1}|=1`
`<=>|-2a^2|=|a-1|` `(a>=0)`
`<=>`\(\left[ \begin{array}{l}a-1=-2a^2\\a-1=-(-2a^2)\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}2a^2+a-1=0\\2a^2-a+1=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}2a^2+a-1=0\\2(a-\dfrac{1}{4})^2+\dfrac{7}{8}=0(VN)\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}a=\dfrac{1}{2}(TM)\\a=-1(KTM)\end{array} \right.\)
Vậy `a=1/2`
