Đáp án + Giải thích các bước giải:
`a//`
`2.|(1)/(2)-(3)/(4)|+\sqrt{(4)/(9)}`
`=2.|-(1)/(4)|+(2)/(3)`
`=2.(1)/(4)+(2)/(3)`
`=(1)/(2)+(2)/(3)`
`=(7)/(6)`
`b//`
`(2x-1)^{7}=(2x-1)^{5}`
`->(2x-1)^{7}-(2x-1)^{5}=0`
`->(2x-1)^{5}[(2x-1)^{2}-1]=0`
`->` \(\left[ \begin{array}{l}(2x-1)^{5}=0\\(2x-1)^{2}-1=0\end{array} \right.\)
`->` \(\left[ \begin{array}{l}2x-1=0\\(2x-1)^{2}=1\end{array} \right.\)
`->` \(\left[ \begin{array}{l}2x=1\\2x-1=±1\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=\dfrac{1}{2}\\2x=0\\2x=2\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=0\\x=1\end{array} \right.\)
Vậy `x∈{(1)/(2);0;1}`
