$(x-5)^2=25$ $TH1: (x-5)^2=(5)^2$ $TH2: (x-5)^2=(-5)^2$ $⇒x-5=5$ $⇒x-5=-5$ $⇒x=5+5$ $⇒x=-5+5$ $⇒x=10$ $⇒x=0$ Vậy `x∈{10;0}` Trả lời
Đáp án: (x – 5)² = 25 ⇔ \(\left[ \begin{array}{l}(x-5)²=5²\\(x-5)²=(-5)²\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x-5=5\\x-5=-5\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x=10\\x=0\end{array} \right.\) Trả lời
$(x-5)^2=25$
$TH1: (x-5)^2=(5)^2$ $TH2: (x-5)^2=(-5)^2$
$⇒x-5=5$ $⇒x-5=-5$
$⇒x=5+5$ $⇒x=-5+5$
$⇒x=10$ $⇒x=0$
Vậy `x∈{10;0}`
Đáp án:
(x – 5)² = 25
⇔ \(\left[ \begin{array}{l}(x-5)²=5²\\(x-5)²=(-5)²\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x-5=5\\x-5=-5\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=10\\x=0\end{array} \right.\)