3.4:
a: |2x-5|=x+1
=>\(\begin{cases}x+1\ge0\\ \left(2x-5\right)^2=\left(x+1\right)^2\end{cases}\Rightarrow\begin{cases}x\ge-1\\ \left(2x-5-x-1\right)\left(2x-5+x+1\right)=0\end{cases}\)
=>\(\begin{cases}x\ge-1\\ \left(x-6\right)\left(3x-4\right)=0\end{cases}\Rightarrow x\in\left\lbrace6;\frac43\right\rbrace\)
b: \(\left|3x-2\right|-1=x\)
=>|3x-2|=x+1
=>\(\begin{cases}x+1\ge0\\ \left(3x-2\right)^2=\left(x+1\right)^2\end{cases}\Rightarrow\begin{cases}x\ge-1\\ \left(3x-2-x-1\right)\left(3x-2+x+1\right)=0\end{cases}\)
=>\(\begin{cases}x\ge-1\\ \left(2x-3\right)\left(4x-1\right)=0\end{cases}\Rightarrow x\in\left\lbrace\frac32;\frac14\right\rbrace\)
c: \(\left|3x-7\right|=2x+1\)
=>\(\begin{cases}2x+1\ge0\\ \left(3x-7\right)^2=\left(2x+1\right)^2\end{cases}\Rightarrow\begin{cases}2x\ge-1\\ \left(3x-7\right)^2-\left(2x+1\right)^2=0\end{cases}\)
=>\(\begin{cases}x\ge-\frac12\\ \left(3x-7-2x-1\right)\left(3x-7+2x+1\right)=0\end{cases}\Rightarrow\begin{cases}x\ge-\frac12\\ \left(x-8\right)\left(5x-1\right)=0\end{cases}\)
=>\(x\in\left\lbrace8;\frac15\right\rbrace\)
d: \(\left|2x-1\right|+1=x\)
=>|2x-1|=x-1
=>\(\begin{cases}x-1\ge0\\ \left(2x-1\right)^2=\left(x-1\right)^2\end{cases}\Rightarrow\begin{cases}x\ge1\\ \left(2x-1-x+1\right)\left(2x-1+x-1\right)=0\end{cases}\)
=>\(\begin{cases}x\ge1\\ x\left(3x-2\right)=0\end{cases}\Rightarrow x\in\) ∅
Bài 3.1:
a: \(\left|\frac12x\right|=3-2x\)
=>|x|=6-4x
=>\(\begin{cases}6-4x\ge0\\ \left(6-4x\right)^2=x^2\end{cases}\Rightarrow\begin{cases}4x\le6\\ \left(6-4x-x\right)\left(6-4x+x\right)=0\end{cases}\)
=>\(\begin{cases}x\le\frac32\\ \left(6-5x\right)\left(6-3x\right)=0\end{cases}\Rightarrow x\in\left\lbrace\frac65\right\rbrace\)
b: \(\left|x-1\right|=3x+2\)
=>\(\begin{cases}3x+2\ge0\\ \left(3x+2\right)^2=\left(x-1\right)^2\end{cases}\Rightarrow\begin{cases}3x\ge-2\\ \left(3x+2-x+1\right)\left(3x+2+x-1\right)=0\end{cases}\)
=>\(\begin{cases}x\ge-\frac23\\ \left(2x+3\right)\left(4x+1\right)=0\end{cases}\Rightarrow x=-\frac14\)
c: \(\left|5x\right|=x-12\)
=>\(\begin{cases}x-12\ge0\\ \left(5x\right)^2=\left(x-12\right)^2\end{cases}\Rightarrow\begin{cases}x\ge12\\ \left(5x-x+12\right)\left(5x+x-12\right)=0\end{cases}\)
=>\(\begin{cases}x\ge12\\ \left(4x+12\right)\left(6x-12\right)=0\end{cases}\Rightarrow x\in\) ∅
d: \(\left|7-x\right|=5x+1\)
=>|x-7|=5x+1
=>\(\begin{cases}5x+1\ge0\\ \left(5x+1\right)^2=\left(x-7\right)^2\end{cases}\Rightarrow\begin{cases}5x\ge-1\\ \left(5x+1-x+7\right)\left(5x+1+x-7\right)=0\end{cases}\)
=>\(\begin{cases}x\ge-\frac15\\ \left(4x+8\right)\left(6x-6\right)=0\end{cases}\Rightarrow x=1\)
