Đáp án + Giải thích các bước giải:
`a)`
`(x-(2)/(3))^{2}=(9)/(4)`
`->` \(\left[ \begin{array}{l}x-\dfrac{2}{3}=\dfrac{3}{2}\\x-\dfrac{2}{3}=-\dfrac{3}{2}\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=\dfrac{13}{6}\\x=-\dfrac{5}{6}\end{array} \right.\)
`b)`
`((4)/(3))^{x}=((16)/(9))^{2}`
`->((4)/(3))^{x}=[((4)/(3))^{2}]^{2}`
`->((4)/(3))^{x}=((4)/(3))^{4}`
`->x=4`
`c)`
`((2)/(3))^{x+1}=(16)/(81)`
`->((2)/(3))^{x+1}=((2)/(3))^{4}`
`->x+1=4`
`->x=3`
`d)`
`5^{(x-2)(x+1)}=1`
`->5^{(x-2)(x+1)}=5^{0}`
`->(x-2)(x+1)=0`
`->` \(\left[ \begin{array}{l}x=2\\x=-1\end{array} \right.\)
`e)`
`(x-1)^{15}=(x-1)^{13}`
`->(x-1)^{15}-(x-1)^{13}=0`
`->(x-1)^{13}[(x-1)^{2}-1]=0`
`->` \(\left[ \begin{array}{l}x-1=0\\(x-1)^{2}-1=0\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=1\\x-1=±1\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=1\\x=2\\x=0\end{array} \right.\)
Vậy `x∈{0;1;2}`
