Lời giải:
`a)`
`(x+2)^(2)=4`
`⇒(x+2)^2=(±2)^2`
$⇒\left[\begin{matrix}x+2=2\\ x+2=-2\end{matrix}\right.$
$⇒\left[\begin{matrix}x=0\\ x=-4\end{matrix}\right.$
Vậy `x∈{0;-4}`
`b)`
`(x-1/4)^3=1/125`
`⇒(x-1/4)^3=(1/5)^3`
`⇒x-1/4=1/5`
`⇒x=1/5+1/4`
`⇒x=9/20`
Vậy `x=9/20`
`c)`
`(x+9)^2=1`
`⇒(x+9)^2=(±1)^2`
$⇒\left[\begin{matrix}x+9=1\\ x+9=-1\end{matrix}\right.$
$⇒\left[\begin{matrix}x=-8\\ x=-10\end{matrix}\right.$
Vậy `x∈{-8;-10}`
`d)`
`(x+1/3)^2=0`
`⇒x+1/3=0`
`⇒x=-1/3`
Vậy `x=-1/3`
`a.`
`(x-1/2)^2=0`
`⇒x-1/2=0`
`⇒x=1/2`
Vậy `x=1/2`
`b.`
`(x-2)^2=1`
`⇒(x-2)^2=(±1)^2`
$⇒\left[\begin{matrix}x-2=1\\ x-2=-1\end{matrix}\right.$
$⇒\left[\begin{matrix}x=3\\ x=1\end{matrix}\right.$
Vậy `x∈{3;1}`
`c.`
`(2x-1)^3=-8`
`⇒(2x-1)^3=(-2)^3`
`⇒2x-1=-2`
`⇒2x=-1`
`⇒x=-1/2`
Vậy `x=-1/2`
`d.`
`(x+1/2)^2=1/16`
`⇒(x+1/2)^2=(±1/4)^2`
$⇒\left[\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{4}\\ x+\dfrac{1}{2}=\dfrac{-1}{4}\end{matrix}\right.$
$⇒\left[\begin{matrix}x=\dfrac{-1}{4}\\ x=\dfrac{-3}{4}\end{matrix}\right.$
Vậy `x∈{-1/4;-3/4}`
`e)`
`3^(x-11/13)=1`
`⇔3^(x-11/13)=3^0`
`⇔x-11/13=0`
`⇔x=11/13`
Vậy `x=11/13`
