I don't know

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

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

\(\left|2x-1\right|=0\)

\(\Leftrightarrow2x-1=0\Leftrightarrow x=\frac{1}{2}\)

KL:....................

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

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

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

KL:........................................

Câu 3 tương tự

lát mk làm tiếp cho

Ta có: \(\hept{\begin{cases}\left|x^2-9\right|\ge0\forall x\\\left|x+3\right|\ge0\forall x\end{cases}}\)

Mà \(\left|x^2-9\right|+\left|x+3\right|=0\)

\(\Rightarrow\hept{\begin{cases}\left|x^2-9\right|=0\\\left|x+3\right|=0\end{cases}\Leftrightarrow\hept{\begin{cases}x^2-9=0\\x=-3\end{cases}\Leftrightarrow}\hept{\begin{cases}x^2=9\\x=-3\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\pm3\\x=-3\end{cases}\Rightarrow}x=-3}\)

Vậy \(x=-3\)

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

\(\Rightarrow x-2\ge0\forall x\)

\(\Rightarrow x\ge2\)

Vậy \(x\ge2\)

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

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

\(\Rightarrow x-3\le0\)

\(\Rightarrow x\le3\)

Vậy \(x\le3\)

a: =>|2x+3|=2+2x-5=2x-3

\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{3}{2}\\\left(2x-3-2x-3\right)\left(2x-3+2x+3\right)=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

b: Trường hợp 1: x<-3

Pt sẽ là -x-3+1-2x=10

=>-3x-2=10

=>-3x=8

hay x=-8/3(loại)

Trường hợp 2: -3<=x<1/2

Pt sẽ la x+3+1-2x=10

=>4-x=10

hay x=-6(loại)

Trường hợp 3: x>=1/2

Pt sẽ là x+3+2x-1=10

=>3x+2=10

hay x=8/3(nhận)

\(c)\) \(\left|2x-1\right|-2x=3\)

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

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 ~ 

Bài 4: 

b: Ta có: \(2x\left(x-\dfrac{1}{4}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{4}\end{matrix}\right.\)