a)
Ta có: |3x|=x+7(1)
Trường hợp 1: x≥0
(1)⇔3x=x+7
⇔3x-x-7=0
⇔2x-7=0
⇔2x=7
⇔\(x=\frac{7}{2}\)(tm)
Trường hợp 2: x<0
(1)⇔-3x=x+7
⇔-3x-x-7=0
⇔-4x-7=0
⇔-4x=7
hay \(x=\frac{7}{-4}=-\frac{7}{4}\)(tm)
Vậy: \(S=\left\{\frac{7}{2};-\frac{7}{4}\right\}\)
b)
Ta có: |-4x|=-2x+11(2)
Trường hợp 1: x≤0
(2)⇔-4x=-2x+11
⇔-4x+2x-11=0
⇔-2x-11=0
⇔-2x=11
hay \(x=-\frac{11}{2}\)(tm)
Trường hợp 2: x>0
(2)⇔4x=-2x+11
⇔4x+2x-11=0
⇔6x-11=0
⇔6x=11
hay \(x=\frac{11}{6}\)(tm)
Vậy: \(S=\left\{-\frac{11}{2};\frac{11}{6}\right\}\)
c) Ta có: |x-9|=2x+5(3)
Trường hợp 1: x≥9
\(\left(3\right)\Leftrightarrow x-9=2x+5\)
\(\Leftrightarrow x-9-2x-5=0\)
\(\Leftrightarrow-x-14=0\)
\(\Leftrightarrow-x=14\)
hay x=-14(loại)
Trường hợp 2: x<9
\(\left(3\right)\Leftrightarrow9-x=2x+5\)
\(\Leftrightarrow9-x-2x-5=0\)
\(\Leftrightarrow4-3x=0\)
\(\Leftrightarrow3x=4\)
hay \(x=\frac{4}{3}\)(tm)
Vậy: \(S=\left\{\frac{4}{3}\right\}\)
d) Ta có: \(\left|3x\right|-1=4x+1\)
\(\Leftrightarrow\left|3x\right|=4x+1+1=4x+2\)(4)
Trường hợp 1: x≥0
\(\left(4\right)\Leftrightarrow3x=4x+2\)
\(\Leftrightarrow3x-4x-2=0\)
\(\Leftrightarrow-x-2=0\)
\(\Leftrightarrow-x=2\)
hay x=-2(loại)
Trường hợp 2: x<0
\(\left(4\right)\Leftrightarrow-3x=4x+2\)
\(\Leftrightarrow-3x-4x-2=0\)
\(\Leftrightarrow-7x=2\)
hay \(x=-\frac{2}{7}\)(tm)
Vậy: \(S=\left\{-\frac{2}{7}\right\}\)
