Question

Tìm x biết \(\)

a, \(\left|3x-5\right|-\left|x+2\right|\)=0

b,\(\left|x\right|+\left|x+2\right|=3\)

B nào lm giúp mk vs mk cần gấp mai mk hk r

Thankf b đõ n`

Answer 1

a, \(\left|3x-5\right|-\left|x+2\right|=0\)

\(\Rightarrow\left|3x-5\right|=\left|x+2\right|\)

\(\Rightarrow\left[{}\begin{matrix}3x-5=x+2\\3x-5=-x-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=7\\4x=3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\)

Vậy \(x=\dfrac{7}{2}\) hoặc \(x=\dfrac{3}{4}\)

b, \(\left|x\right|+\left|x+2\right|=3\)

+) Xét \(x\ge0\) ta có:

\(x+x+2=3\Rightarrow2x=1\Rightarrow x=\dfrac{1}{2}\) ( t/m )

+) Xét \(-2\le x< 0\) ta có:
\(-x+x+2=3\Rightarrow2=3\) ( vô lí )

+) Xét \(x< -2\) ta có:

\(-x-x-2=3\Rightarrow-2x=5\Rightarrow x=\dfrac{-5}{2}\) ( t/m )

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

Answer 2

a) \(\left|3x-5\right|-\left|x+2\right|=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}3x-5=x+2\\3x-5=-\left(x+2\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\)

Vậy \(x_1=\dfrac{3}{4};x_2=\dfrac{7}{2}\)

b) \(\left|x\right|+\left|x+2\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}x+x+2=3\left(đk:x\ge0,x+2\ge0\right)\\-x+x+2=3\left(đk:x< 0,x+2\ge0\right)\\x-\left(x+2\right)=3\left(đk:x\ge0,x+2< 0\right)\\-x-\left(x+2\right)=3\left(đk:x< 0,x+2< 0\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(đk:x\ge0,x\ge-2\right)\\x\in\varnothing\left(đk:x< 0,x\ge-2\right)\\x\in\varnothing\left(x\ge0,x< -2\right)\\x=-\dfrac{5}{2}\left(đk:x< 0,x< -2\right)\end{matrix}\right.\)

Vậy \(x_1=-\dfrac{5}{2};x_2=\dfrac{1}{2}\)

Answer 3

Nguyễn Huy Tú tl hộ đc k

Answer 4

toán cô thoa cho hả bảo sao thấy nhiều người đăng câu hỏi thế