Question

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Answer 1

a) \(\sqrt{4\left(1-2x\right)^2}=6\)

\(\Leftrightarrow\sqrt{\left[2\left(1-2x\right)\right]^2}=6\)

\(\Leftrightarrow\left|2\left(1-2x\right)\right|=6\)

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

\(\Leftrightarrow\left[{}\begin{matrix}1-2x=3\\1-2x=-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=-2\\2x=4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\)

b) \(\sqrt{\left(2x-1\right)^2}+\sqrt{9\left(2x-1\right)^2}=8\)

\(\Leftrightarrow\left|2x-1\right|+3\left|2x-1\right|=8\)

\(\Leftrightarrow4\left|2x-1\right|=8\)

\(\Leftrightarrow\left|2x-1\right|=2\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=2\left(x\ge\dfrac{1}{2}\right)\\2x-1=-2\left(x< \dfrac{1}{2}\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=3\\2x=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\left(tm\right)\)

c) \(\sqrt{4x^2+4x+1}=5\)

\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=5\)

\(\Leftrightarrow\left|2x+1\right|=5\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=5\left(x\ge-\dfrac{1}{2}\right)\\2x+1=-5\left(x< -\dfrac{1}{2}\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=4\\2x=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\left(tm\right)\) 

Mình làm mẫu vài câu thôi nhé do dạng bài này rất dài !