Question

Tìm x

a) \(\sqrt{\left(x-3\right)^2}=9\)

b)\(\sqrt{4x^2+4x+1}=6\)

Answer 1

\(a,\sqrt{\left(x-3\right)^2}=9\)

=> \(\left|x-3\right|=9\)

=> \(\left[{}\begin{matrix}x-3=9\\x-3=-9\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=9+3=12\\x=-9+3=-6\end{matrix}\right.\)

Vậy \(x\in\left\{12;-6\right\}\)

=> \(\sqrt{\left(2x+1\right)^2}=6\) ( phân tích đa thức thành nhân tử )

=> \(\left|2x+1\right|=6\)

=> \(\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}2x=6-1=5\\2x=-6-1=-7\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=5:2=\frac{5}{2}\\x=-7:2=-\frac{7}{2}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{5}{2};-\frac{7}{2}\right\}\)