\(\left|2x+3\right|-3=3\)
<=> \(\left|2x+3\right|=6\)
Ta có : \(\left|2x+3\right|=\left\{{}\begin{matrix}2x+3khix\ge\frac{-3}{2}\\-\left(2x+3\right)khix\le\frac{-3}{2}\end{matrix}\right.\)
Ta giải 2 phương trình
1) \(\left|2x+3\right|=6\) khi x ≥ \(\frac{-3}{2}\)
<=> 2x+3=6
<=> x= \(\frac{3}{2}\)
2) \(\left|2x+3\right|=6\) khi x ≤ \(\frac{-3}{2}\)
<=> -(2x+3)=6
<=> -2x-3=6
<=>x= \(\frac{-9}{2}\)
S những bài lớp lớn mà bn cứ đăng ở lớp 6 thế!!!
\(\left|2x+3\right|-2=3\)
\(\Leftrightarrow\left|2x+3\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=5khi2x+3\ge0\\-\left(2x+3\right)=5khi2x+3< 0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=2khix\ge-\frac{3}{2}\\-2x-3=5khix< -\frac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1khix\ge-\frac{3}{2}\left(tm\right)\\x=-4khix< -\frac{3}{2}\left(tm\right)\end{matrix}\right.\)
Vậy x\(\in\left\{1;-4\right\}\)
\(\left|2x+3\right|-2=3\\ \left|2x+3\right|=3+2\\ \left|2x+3\right|=5\\ \Rightarrow\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)
vậy x={1;-4}
/2.x+3/-2=3
/2.x+3/=3+2
/2.x+3/=5
=>2.x+3=5 hoặc 2.x+3=(-5)
*2.x+3=5
2.x=5-3
2.x=2
x=2:2
x=1
*2.x+3=(-5)
2.x=(-5)-3
2.x=(-8)
x=(-8):2
x=(-4)
Vậy x= (-4) hoặc x=1