ĐKXĐ: \(x\ge-15\)
\(\sqrt{16\left(x-1\right)^2}=2x+\sqrt{12}\cdot\sqrt{75}\)
\(\Leftrightarrow\sqrt{16\left(x-1\right)^2}=2x+30\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-4=2x+30\\4x-4=-2x-30\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=17\\x=-\frac{13}{3}\end{matrix}\right.\)(Thỏa mãn)
Vậy phương trình có tập nghiệm \(S=\left\{17;-\frac{13}{3}\right\}\)
ĐKXĐ: \(x\ge-15\)
Ta có: \(\sqrt{16\left(x-1\right)^2}=2x+\sqrt{12}\cdot\sqrt{75}\)
\(\Leftrightarrow\sqrt{\left[4\left(x-1\right)\right]^2}=2x+30\)
\(\Leftrightarrow\left|4x-4\right|=2x+30\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-4=2x+30\\4x-4=-2x-30\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x-4-2x-30=0\\4x-4+2x+30=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x-34=0\\6x+26=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=34\\6x=-26\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=17\left(nhận\right)\\x=-\frac{26}{6}\left(nhận\right)\end{matrix}\right.\)
Vậy: \(x\in\left\{17;\frac{-26}{6}\right\}\)
\(\sqrt{16\left(x-1\right)^2}=2x+\sqrt{12}.\sqrt{75}\)
\(\Leftrightarrow\)\(4\sqrt{\left(x-1\right)^2}=2x+\sqrt{12.75}\)
\(\Leftrightarrow4x-4=2x+\sqrt{900}\)
\(\Leftrightarrow4x-2x=4+30\) (chuyển vế đổi dấu)
\(\Leftrightarrow2x=34\)
\(\Leftrightarrow x=17\)
