a, \(\sqrt{\left(x-3\right)^2}=9\)
\(\Leftrightarrow\left|x-3\right|=9\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=9\\x-3=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-6\end{matrix}\right.\)
Vậy...
b, \(\sqrt{4x^2+4x+1}=6\)
\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)
\(\Leftrightarrow\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
Vậy...
a) \(\sqrt{\left(x-3\right)^2}=9\) (1)
\(\Leftrightarrow\left|x-3\right|=9\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=9\\x-3=-9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-6\end{matrix}\right.\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{-6;12\right\}\)
b) \(\sqrt{4x^2+4x+1}=6\) (2)
\(\Leftrightarrow4x^2+4x+1=36\)
\(\Leftrightarrow\left(2x+1\right)^2=36\)
\(\Leftrightarrow2x+1=\pm6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=-7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
Vậy tập nghiệm phương trình (2) là \(S=\left\{-\dfrac{7}{2};\dfrac{5}{2}\right\}\)
\(\sqrt{\left(x-3\right)^2}=9\)
\(\Leftrightarrow\left|x-3\right|=9\)
\(\Rightarrow\left[{}\begin{matrix}x-3=9\\x-3=-9\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=12\\x=-6\end{matrix}\right.\)
Vậy..
b) \(\sqrt{4x^2+4x+1}=6\)
\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)
\(\Leftrightarrow\left|2x+1\right|=6\)
\(\Rightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-7}{2}\end{matrix}\right.\)
Vậy..
