Đáp án và giải thích các bước giải:
`9B`
`a)` `\sqrt[2x^2-2x+1]=2x-1`
`⇔` `2x^2-2x+1=(2x-1)^2`
`⇔` `2x^2-2x+1=4x^2-4x+1`
`⇔` `2x^2-2x=0`
`⇔` `2x(x-1)=0`
`⇔` \(\left[ \begin{array}{l}2x=0\\x-1=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=0\\x=1\end{array} \right.\)
Vậy `S={0;1}`
`b)` `\sqrt[x+4\sqrt[x-4]]=2` `(ĐK: x≥4)`
`⇔` `x+4\sqrt[x-4]=4`
`⇔` `4\sqrt[x-4]=4-x`
`⇔` `16(x-4)=(4-x)^2`
`⇔` `16x-64=16-8x+x^2`
`⇔` `x^2-24x+80=0`
`⇔` `(x-20)(x-4)=0`
`⇔` \(\left[ \begin{array}{l}x-20=0\\x-4=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=20(TMĐK)\\x=4(TMĐK)\end{array} \right.\)
Vậy `S={20;4}`
`10A`
`a)` `\sqrt[x^2-3x+2]=\sqrt[x-1]` `(ĐK:x≥1)`
`⇔` `x^2-3x+2=x-1`
`⇔` `x^2-4x+3=0`
`⇔` `(x-3)(x-1)=0`
`⇔` \(\left[ \begin{array}{l}x-3=0\\x-1=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=3(TMĐK)\\x=1(TMĐK)\end{array} \right.\)
Vậy `S={3;1}`
`b)` `\sqrt[x^2-4x+4]=\sqrt[4x^2-12x+9]`
`⇔` `\sqrt[(x-2)^2]=\sqrt[(2x-3)^2]`
`⇔` `|x-2|=|2x-3|`
`⇔` \(\left[ \begin{array}{l}x-2=2x-3\\x-2=3-2x\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=1\\x=\dfrac53\end{array} \right.\)
Vậy `S={1;5/3}`
