Question

giải các phương trình sau:

a)\(\left|4x+7\right|=2x+5\) b)\(\left|x^2-4x-5\right|=\left|4x-17\right|\)

c)\(\left|2x-5\right|+\left|2x^2-7x+5\right|\)=0 d)\(\left|x-1\right|+\left|2x+1\right|=\left|3x\right|\)

Answer 1

a/ \(x\ge-\frac{5}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+7=2x+5\\4x+7=-2x-5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\end{matrix}\right.\)

b/ \(\Leftrightarrow\left[{}\begin{matrix}x^2-4x-5=4x-17\\x^2-4x-5=17-4x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-8x+12=0\\x^2=22\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2\\x=6\\x=\pm\sqrt{22}\end{matrix}\right.\)

c/ \(\Leftrightarrow\left\{{}\begin{matrix}2x-5=0\\2x^2-7x+5=0\end{matrix}\right.\) \(\Rightarrow x=\frac{5}{2}\)

d/ \(\left|x-1\right|+\left|2x+1\right|\ge\left|x-1+2x+1\right|=\left|3x\right|\)

Dấu "=" xảy ra khi và chỉ khi: \(\left(x-1\right)\left(2x+1\right)\ge0\Leftrightarrow\left[{}\begin{matrix}x\le-\frac{1}{2}\\x\ge1\end{matrix}\right.\)

Vậy nghiệm của pt là \(\left[{}\begin{matrix}x\le-\frac{1}{2}\\x\ge1\end{matrix}\right.\)