Question

3.Giải các phương trình sau a)\(|x-7|=2x+3\)

Answer 1

\(\left|x-7\right|=2x+3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-7=2x+3\left(ĐK:x-7\ge0\right)\\7-x=2x+3\left(Đk:x-7< 0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2x=3+7\left(ĐK:x\ge7\right)\\-x-2x=3-7\left(ĐK:x< 7\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-x=10\left(ĐK:x\ge7\right)\\-3x=-4\left(ĐK:x< 7\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-10\\x=\dfrac{4}{3}\end{matrix}\right.\)( x = -10 loại )

Vậy pt trên có nghiệm \(x=\dfrac{4}{3}\)

Answer 2

Ta có:

\(\left|x-7\right|=2x+3\) ( ĐK: x \(\ge-\dfrac{3}{2}\))

\(\Leftrightarrow\left[{}\begin{matrix}x-7=2x+3\\x-7=-2x-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2x=3+7\\x+2x=-3+7\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-10\left(l\right)\\x=\dfrac{4}{3}\left(n\right)\end{matrix}\right.\)

Vậy tập nghiệm của pt là: \(S=\left\{\dfrac{4}{3}\right\}\)

Answer 3

Đây là bài làm của mình

Answer 4

\(\left|x-7\right|=2x+3\)(\(đk:x\ge-\dfrac{3}{2}\))

\(\Rightarrow\left(\left|x-7\right|\right)^2=\left(2x+3\right)^2\)

\(\Leftrightarrow\left(x-7\right)^2=\left(2x+3\right)^2\)

\(\Leftrightarrow\left(x-7\right)^2-\left(2x+3\right)^2=0\)

\(\Leftrightarrow\left(-x-10\right)\left(3x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-x-10=0\\3x-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-10\left(loai\right)\\x=\dfrac{4}{3}\left(tm\right)\end{matrix}\right.\)