Question

[Nguyễn Linh]

Bài 2. Giải các phương trình sau:
a) |x-2|+2x=7
b) |x-3| -4x=5

c) |2x+3|+x=2x+3
d) |x+2|=|3x-4|

Answer 1

[chuche]

a, \(x<2\)

\(2-x+2x=7\)

\(x=5(\)ko \(t/m)\)

\(x>2\)

\(-x=5\)

\(x=-5(ko\) \(t/m)\)

Answer 2

[Nguyễn Lê Phước Thịnh]

a: |x-2|+2x=7

=>|x-2|=-2x+7

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{7}{2}\\\left(-2x+7\right)^2=\left(x-2\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{7}{2}\\\left(2x-7-x+2\right)\left(2x-7+x-2\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{7}{2}\\\left(x-5\right)\left(3x-9\right)=0\end{matrix}\right.\Leftrightarrow x=3\)

b: |x-3|-4x=5

=>|x-3|=4x+5

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{5}{4}\\\left(4x+5-x+3\right)\left(4x+5+x-3\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{5}{4}\\\left(3x+8\right)\left(5x+2\right)=0\end{matrix}\right.\Leftrightarrow x=-\dfrac{2}{5}\)

c: |2x+3|+x=2x+3

=>|2x+3|=x+3

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-3\\\left(2x+3-x-3\right)\left(2x+3+x+3\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{0;-2\right\}\)