Bài 1:
a: \(\left|x-\frac43\right|=2\)
=>\(\left[\begin{array}{l}x-\frac43=2\\ x-\frac43=-2\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2+\frac43=\frac{10}{3}\\ x=-2+\frac43=-\frac23\end{array}\right.\)
b: \(\left|\frac52-x\right|=\frac14\)
=>\(\left|x-\frac52\right|=\frac14\)
=>\(\left[\begin{array}{l}x-\frac52=\frac14\\ x-\frac52=-\frac14\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac52+\frac14=\frac{11}{4}\\ x=\frac52-\frac14=\frac94\end{array}\right.\)
c: \(\left|\frac43x-1\right|-\frac12=\frac32\)
=>\(\left|\frac43x-1\right|=\frac32+\frac12=2\)
=>\(\left[\begin{array}{l}\frac43x-1=2\\ \frac43x-1=-2\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac43x=2+1=3\\ \frac43x=-2+1=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3:\frac43=\frac94\\ x=-1:\frac43=-\frac34\end{array}\right.\)
d: \(\left|\frac23x-4\right|=\left|5-\frac72\right|\)
=>\(\left|\frac23x-4\right|=\frac32\)
=>\(\left[\begin{array}{l}\frac23x-4=\frac32\\ \frac23x-4=-\frac32\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac23x=4+\frac32=\frac{11}{2}\\ \frac23x=-\frac32+4=\frac52\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{11}{2}:\frac23=\frac{11}{2}\cdot\frac32=\frac{33}{4}\\ x=\frac52:\frac23=\frac52\cdot\frac32=\frac{15}{4}\end{array}\right.\)
e: |3x+2|=|5-2x|
=>3x+2=5-2x hoặc 3x+2=2x-5
=>5x=3 hoặc 3x-2x=-5-2
=>x=3/5 hoặc x=-7
Bài 2:
a: \(\left|2x-\frac14\right|\ge0\forall x\)
=>\(\left|2x-\frac14\right|+1\ge1\forall x\)
Dấu '=' xảy ra khi \(2x-\frac14=0\)
=>\(2x=\frac14\)
=>x=1/8
b: \(\left|5-\frac32x\right|\ge0\forall x\)
=>\(\left|5-\frac32x\right|-3\ge-3\forall x\)
Dấu '=' xảy ra khi \(5-\frac32x=0\)
=>\(\frac32x=5\)
=>\(x=5:\frac32=5\cdot\frac23=\frac{10}{3}\)
c: \(\left|x-2020\right|\ge0\forall x\)
\(\left|y-2022\right|\ge0\forall y\)
Do đó: \(\left|x-2020\right|+\left|y-2022\right|\ge0\forall x,y\)
=>\(\left|x-2020\right|+\left|y-2022\right|+1\ge1\forall x,y\)
Dấu '=' xảy ra khi x-2020=0 và y-2022=0
=>x=2020 và y=2022
