Đáp án:
Giải thích các bước giải:
a) `(x-3)^2=9`
`⇔ (x-3)^2=(3)^2`
`⇔` \(\left[ \begin{array}{l}x-3=3\\x-3=-3\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=6\\x=0\end{array} \right.\)
Vậy `x \in {0;6}`
b) `(x+6)^2=25`
`⇔ (x+6)^2=5^2`
`⇔` \(\left[ \begin{array}{l}x+6=5\\x+6=-5\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=-1\\x=-11\end{array} \right.\)
Vậy `x \in {-1;-11}`
c) `(y-7)^2-4=32
`⇔ (y-7)^2=36`
`⇔ (y-7)^2=(6)^2`
`⇔` \(\left[ \begin{array}{l}y-7=6\\y-7=-6\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}y=13\\y=1\end{array} \right.\)
Vậy `y \in {1;13}`
d) `15-(y+2)=14`
`⇔ 15-14=y+2`
`⇔ 1=y+2`
`⇔ y=-2+1`
`⇔ y=-1`
Vậy `y \in {-1}`
