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

cha biet

\(||3x-1|-\dfrac{1}{2}|=\dfrac{5}{2}\)

Có thể xảy ra 2 trường hợp: 

TH1:\(||3x-1|-\dfrac{1}{2}|=-\dfrac{5}{2}\)

TH2: \(||3x-1|-\dfrac{1}{2}|=\dfrac{5}{2}\)

Giả sử \(|3x-1|-\dfrac{1}{2}=-\dfrac{5}{2}\)

⇔    \(|3x-1|=-\dfrac{5}{2}+\dfrac{1}{2}\)

⇔    \(|3x-1|=-2\) (Vô lí, vì |3x - 1| ≥ 0 ∀ x)

⇒ \(|3x-1|-\dfrac{1}{2}=\dfrac{5}{2}\)

⇔ \(|3x-1|=\dfrac{5}{2}+\dfrac{1}{2}\)

⇔ \(|3x-1|=3\)

⇔ \(3x-1\in\left\{\pm3\right\}\)

⇔ \(3x\in\left\{-2;4\right\}\)

⇔ \(x\in\left\{-\dfrac{2}{3};\dfrac{4}{3}\right\}\)

Vậy \(x\in\left\{-\dfrac{2}{3};\dfrac{4}{3}\right\}\)

\(\left|\left|3x-1\right|-\dfrac{1}{2}\right|=\dfrac{5}{2}\)

\(\Rightarrow\)2 trường hợp:

Th1:\(3x-1-\dfrac{1}{2}=\dfrac{5}{2}\)

\(3x-1=\dfrac{5}{2}+\dfrac{1}{2}\)

\(3x-1=3\)

\(3x=3+1\)

\(3x=4\Rightarrow x=4:3\Rightarrow x=\dfrac{4}{3}\)

Th2:

\(3x-1-\dfrac{1}{2}=-\dfrac{5}{2}\)

\(3x-1=-\dfrac{5}{2}+\dfrac{1}{2}\)

\(3x-1=-2\)

\(3x=-2+1\)

\(3x=-1\Rightarrow x=-1:3\Rightarrow x=\dfrac{-1}{3}\)

P/s Mình làm theo cách chửa mình nếu sai thì xin lỗi bạn nha

a: =>2x+5=x-1 hoặc 2x+5=-x+1

=>x=-6 hoặc 3x=-4

=>x=-6 hoặc x=-4/3

b: =>2x+7=x-3 hoặc 2x+7=-x+3

=>x=-10 hoặc 3x=-4

=>x=-10 hoặc x=-4/3

c: =>3x-2=2x+3 hoặc 3x-2=-2x-3

=>x=5 hoặc 5x=-1

=>x=5 hoặc x=-1/5

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

\(\Rightarrow\left|x-1\right|=1-3x\)

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

\(\Rightarrow\orbr{\begin{cases}x+\left(-1\right)=1+\left(-3x\right)\\x+\left(-1\right)=1+3x\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x-\left(-3x\right)=1-\left(-1\right)\\x-3x=1-\left(-1\right)\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}4x=2\\-2x=2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-1\end{cases}}\)

A ) | x - 1 | + 3x = 1

        x - 1 + 3x   = 1

     ( x + 3x ) - 1 = 1

       4x         - 1  = 1

               4x       = 1 + 1

               4x       = 2

                 x        = 2 : 4

                 x        = 0,5

Vậy x = 0,5

B) | 5x - 3 | - x = 7

       5x - 3 - x   = 7

    ( 5x - x ) - 3 = 7

        4x       - 3 = 7

                 4x   = 7 + 3

                 4x   = 10

                   x    = 10 : 4

                   x    = 2,5

Vậy x = 2,5 

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

\(\Rightarrow\left|5x-2\right|=7+x\)

\(\Rightarrow\orbr{\begin{cases}5x-2=7-x\\5x-2=7+x\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}5x+\left(-2\right)=7+\left(-x\right)\\5x+\left(-2\right)=7+x\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}5x-\left(-x\right)=7-\left(-2\right)\\5x-x=7-\left(-2\right)\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}6x=9\\4x=9\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{9}{4}\end{cases}}\)