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1.
2|x-6|+7x-2=|x-6|+7x
2|x-6| - |x-6|=7x-(7x-2)
|x-6| = 2
=>x-6 = +2
*x-6=2 *x-6 = -2
x =2+6 x = (-2)+6
x =8 x = 4
2.
|x-5|-7(x+4)=5-7x
|x-5|-7x-28 =5-7x
|x-5|-28 =5-7x+7x
|x-5|-28 = 5
|x-5| = 5+28
|x-5| = 33
=>x-5 = +33
*x-5=33 *x-5=-33
x =38 x = -28
3.
3|x+4|-2(x-1)=7-2x
3|x+4|-2x+2 =7-2x
3|x+4|-2 =7-2x+2x
3|x+4|-2 =7
3|x+4| =7+2
3|x+4| = 9
|x+4| =9:3
|x+4| = 3
=>x+4 = +3
*x+4=3 *x+4=-3
x =-1 x = -7
\(2\left|x-1\right|+3\left(x+2\right)=3^2\)
\(2\left|x-1\right|+3x+6=9\)
\(2\left|x-1\right|=9-3x-6\)
\(2\left|x-1\right|=3-3x\)
\(\left|x-1\right|=\frac{3-3x}{2}\)
\(\Rightarrow\orbr{\begin{cases}x-1=\frac{3-3x}{2}\\x-1=-\frac{3-3x}{2}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{3-3x}{2}+\frac{2}{2}\\x=\frac{-3+3x}{2}+\frac{2}{2}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{3-3x+2}{2}\\x=\frac{-3+3x+2}{2}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{5-3x}{2}\\x=\frac{-1+3x}{2}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=5-3x\\2x=-1+3x\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}5x=5\\x=1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=1\end{cases}}\)
\(\frac{-2}{3}\) \(-\) \(\frac{1}{3}\) X \(\left(2.x-5\right)\) \(=\frac{3}{2}\)
\(-1\) X \(\left(2.x-5\right)\) \(=\frac{3}{2}\)
\(\left(2.x-5\right)\) \(=\frac{3}{2}\) \(:-1\)
\(\left(2.x-5\right)\) \(=\frac{3}{2}\)
\(2.x\) \(=\frac{3}{2}\) \(+\) \(5\)
\(2.x\) \(=\frac{7}{2}\)
\(x=\) \(\frac{7}{2}\) \(:2\)
\(x=\frac{7}{4}\)
* Mới lớp 5 nên không chắc, sai thongcam *
#Ninh Nguyễn
\(\frac{-2}{3}-\frac{1}{3}\cdot\left(2x-5\right)=\frac{3}{2}\)
\(\frac{1}{3}\left(2x-5\right)=\frac{-2}{3}-\frac{3}{2}\)
\(2x-5=\frac{-13}{6}:\frac{1}{3}\)
\(2x=\frac{-13}{2}+5\)
\(x=\frac{-3}{2}:2\)
\(x=\frac{-3}{4}\)
a.
\(\left|x+10\right|=15\Rightarrow\orbr{\begin{cases}x+10=15\\x+10=-15\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-25\end{cases}}}\)
b.
\(\left|x-3\right|+5=7\Rightarrow\left|x-3\right|=2\Rightarrow\orbr{\begin{cases}x-3=2\\x-3=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=1\end{cases}}\)
c.
\(\left|x-3\right|+12=6\Rightarrow\left|x-3\right|=-6\Rightarrow x=\Phi\)
Phương trình vô nghiệm
d.
\(\left(2x+4\right)\left(3x-9\right)=0\Rightarrow\orbr{\begin{cases}2x+4=0\\3x-9=0\end{cases}\Rightarrow}\orbr{\begin{cases}2x=-4\\3x=9\end{cases}}\Rightarrow\orbr{\begin{cases}x=-2\\x=3\end{cases}}\)
e.
\(x^2-5x=0\Rightarrow x\left(x-5\right)=0\Rightarrow\orbr{\begin{cases}x=0\\x-5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=5\end{cases}}\)
f.
\(\left(x+3\right)\left(4-2x\right)=70\Rightarrow4x-2x^2+7-6x=70\Rightarrow2x^2+2x+63=0\Rightarrow2\left(x+\frac{1}{2}\right)^2+\frac{123}{2}=0\)(vô lí)
Vậy phương trình vô nghiệm
