a) \(\left|4-x\right|+2x=3\)

\(\Rightarrow\left|4-x\right|=3-2x\)

Nếu \(4-x\ge0\Rightarrow x\ge-4\) thì:
\(4-x=3-2x\)

\(\Rightarrow4-3=-2x+x\)

\(\Rightarrow-x=1\)

\(\Rightarrow x=-1\) ( t/m )

Nếu \(4-x< 0\Rightarrow x< -4\) thì:

\(-\left(4-x\right)=3-2x\)

\(\Rightarrow-4+x=3-2x\)

\(\Rightarrow-4-3=-2x-x\)

\(\Rightarrow-7=-3x\)

\(\Rightarrow x=\frac{7}{3}\) ( loại )

Vậy \(x=-1\)

b) Vì \(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|\ge0\)

nên \(4x\ge0\Rightarrow x\ge0\)

\(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|=4x\)

\(\Rightarrow x+1+x+2+x+3=4x\)

\(\Rightarrow x=6\)

Vậy \(x=6\)

c) \(\left|2x-1\right|=2\)

\(\Rightarrow2x-1=\pm2\)

+) \(2x-1=2\Rightarrow x=\frac{3}{2}\)

+) \(2x-1=-2\Rightarrow x=\frac{-1}{2}\)

Vậy \(x\in\left\{\frac{3}{2};\frac{-1}{2}\right\}\)

d) \(\left|3-2x\right|+\left|4y+5\right|=0\)

\(\Rightarrow\left|3-2x\right|=0\) và \(\left|4y+5\right|=0\)

+) \(\left|3-2x\right|=0\Rightarrow3-2x=0\Rightarrow x=\frac{3}{2}\)

+) \(\left|4y+5\right|=0\Rightarrow4y+5=0\Rightarrow y=\frac{-5}{4}\)

Vậy \(x=\frac{3}{2};y=\frac{-5}{4}\)

e) \(x^2+\left|x-1\right|=x^2+2\)

\(\Rightarrow\left|x-1\right|=2\)

Đến đây làm tương tự phần c để tìm x