a) \(\sqrt{\left(x-3\right)^2}=9\)
\(\Leftrightarrow\sqrt{\left(x-3\right)^2}^2=9^2\)
\(\Leftrightarrow\left(x-3\right)^2=81\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=9\\x-3=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-6\end{matrix}\right.\)
b) \(\sqrt{4x^2+4x+1}=6\)
\(\Leftrightarrow\sqrt{4x^2+4x+1}^2=6^2\)
\(\Leftrightarrow4x^2+4x+1-36=0\)
\(\Leftrightarrow4x^2+4x-35=0\)
\(\Leftrightarrow4\cdot\left(x-\dfrac{5}{2}\right)\cdot\left(x+\dfrac{7}{2}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{5}{2}=0\\x+\dfrac{7}{2}=0\end{matrix}\right.\rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
a, \(\sqrt{\left(x-3\right)^2}=9\)
\(\Leftrightarrow\)I x-3 I=3
<=>\(\left[{}\begin{matrix}x-3=3\\x-3=-3\end{matrix}\right.\)
<=>\(\left[{}\begin{matrix}x=6\\x=0\end{matrix}\right.\)
b, \(\sqrt{4x^2+4x+1}=6\)
<=>\(\sqrt{\left(2x+1\right)^2}=6\)
<=> I 2x+1 I =6
<=>\(\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-7}{2}\end{matrix}\right.\)
