Câu 3:
a: Ta có: \(\sqrt{\left(2x+3\right)^2}=5\)
\(\Leftrightarrow\left|2x+3\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)
b: ta có: \(\sqrt{16x-16}-\sqrt{9x-9}+\sqrt{4x-4}+\sqrt{x-1}=8\)
\(\Leftrightarrow4\sqrt{x-1}-3\sqrt{x-1}+2\sqrt{x-1}+\sqrt{x-1}=8\)
\(\Leftrightarrow4\sqrt{x-1}=8\)
\(\Leftrightarrow x-1=4\)
hay x=5
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