viết lại 1 a) l 1/2xl =3 - 2x

1.a) ĐK : \(3-2x\ge0\forall x\Rightarrow x\le\frac{3}{2}\)

Khi đó :  \(\left|\frac{1}{2}x\right|=3-2x\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x=3-2x\\\frac{1}{2}x=-3+2x\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{5}{2}x=3\\\frac{3}{2}x=3\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{6}{5}\\x=2\end{cases}}\left(tm\right)\)

Vậy \(x\in\left\{\frac{6}{5};2\right\}\)

b) ĐK : \(3x+2\ge0\Rightarrow x\ge\frac{-2}{3}\)

Khi đó : \(\left|x-1\right|=3x+2\Leftrightarrow\orbr{\begin{cases}x-1=3x+2\\x-1=-3x-2\end{cases}}\Rightarrow\orbr{\begin{cases}-2x=3\\4x=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1,5\\x=-0,25\left(tm\right)\end{cases}}\)

Vậy x = -0,25

c) ĐKXĐ : \(x-12\ge0\Rightarrow x\ge12\)

Khi đó |5x| = x - 12

<=> \(\orbr{\begin{cases}5x=x-12\\5x=-x+12\end{cases}}\Rightarrow\orbr{\begin{cases}4x=-12\\6x=12\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=2\end{cases}}\left(\text{loại}\right)\)

Vậy \(x\in\varnothing\)

d) ĐK :  \(5x+1\ge0\Rightarrow x\ge-\frac{1}{5}\)

Khi đó \(\left|17-x\right|=5x+1\Leftrightarrow\orbr{\begin{cases}17-x=5x+1\\17-x=-5x-1\end{cases}}\Rightarrow\orbr{\begin{cases}6x=16\\-4x=18\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{8}{3}\left(tm\right)\\x=-4,5\left(\text{loại}\right)\end{cases}}\)

Vậy x = 8/3 

Tóm lại : Cách làm là 

|f(x)| = g(x)

ĐK : g(x) \(\ge0\)

=> \(\orbr{\begin{cases}f\left(x\right)=-g\left(x\right)\\f\left(x\right)=g\left(x\right)\end{cases}}\)

Bạn tự làm tiếp đi ak

\(c)\) \(\left|2x-1\right|-2x=3\)

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

Ta có : \(\left|2x-1\right|\ge0\)

\(\Rightarrow\)\(2x+3\ge0\)\(\Rightarrow\)\(2x\ge-3\)\(\Rightarrow\)\(x\ge\frac{-3}{2}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}2x-1=2x+3\\2x-1=-2x-3\end{cases}\Leftrightarrow\orbr{\begin{cases}2x-2x=3+1\\2x+2x=-3+1\end{cases}}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}0=4\\4x=-2\end{cases}\Leftrightarrow\orbr{\begin{cases}0=4\left(loai\right)\\x=\frac{-1}{2}\left(tm\right)\end{cases}}}\)

Vậy \(x=\frac{-1}{2}\)

Chúc bạn học tốt ~ 

\(b)\) \(3\left(2x-1\right)-\left|x-5\right|=7\)

\(\Leftrightarrow\)\(3\left(2x-1\right)-7=\left|x-5\right|\)

\(\Leftrightarrow\)\(6x-3-7=\left|x-5\right|\)

\(\Leftrightarrow\)\(\left|x-5\right|=6x-10\)

Ta có : \(\left|x-5\right|\ge0\)

\(\Rightarrow\)\(6x-10\ge0\)\(\Rightarrow\)\(6x\ge10\)\(\Rightarrow\)\(x\ge\frac{5}{3}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x-5=6x-10\\x-5=10-6x\end{cases}\Leftrightarrow\orbr{\begin{cases}6x-x=-5+10\\x+6x=10+5\end{cases}}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}5x=5\\7x=15\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\left(loai\right)\\x=\frac{15}{7}\left(tm\right)\end{cases}}}\)

Vậy \(x=\frac{15}{7}\)

Chúc bạn học tốt ~ 

\(\left|2-x\right|+\left|x+1\right|=5\)

TH1 : \(\left|2-x\right|=\pm5\)

+ ) \(2-x=5\)

            \(x=2-5\)

           \(x=-3\)

+ ) \(2-x=\left(-5\right)\)

           \(x=2-\left(-5\right)\)

          \(x=7\)

TH2 : \(\left|x+1\right|=\pm5\)

+ ) \(x+1=5\)

             \(x=5-1\)

            \(x=4\)

+ ) \(x+1=\left(-5\right)\)

       \(x=\left(-5\right)-1\)

      \(x=-6\)

2 ) \(\left|x+1\right|+\left|2x+1\right|=22\)

TH1 : \(\left|x+1\right|=\pm22\)

+ ) \(x+1=22\)

      \(x=22-1\)

      \(x=21\)

+ ) \(x+1=-22\)

      \(x=-22-1\)

     \(x=-23\)

 TH2: \(\left|2x+1\right|=\pm22\)

+ ) \(2x+1=22\)

     \(2x=21\)

      \(x=\frac{21}{2}\)

+ ) \(2x+1=-22\)

     \(2x=-23\)

     \(x=\frac{-23}{2}\)