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\(2\left(x-1\right)+3\left(3x-2\right)=x-4\)
\(2x-2+9x-6=x-4\)
\(2x+9x-x-2-6=-4\)
\(10x-2-6=-4\)
\(10x-2=2\)
\(10x=4\)
\(x=\frac{2}{5}\)
Vậy \(x=\frac{2}{5}\)
\(3\left(4-x\right)-2\left(x-1\right)=x+20\)
\(12-3x-2x+2=x+20\)
\(12-5x+2=x+20\)
\(12-5x-x+2=20\)
\(12-6x+2=20\)
\(12-6x=18\)
\(6x=-6\)
\(x=-1\)
Vậy \(x=-1.\)
\(4\left(2x+7\right)-3\left(3x-2\right)=24\)
\(8x+28-9x+6=24\)
\(8x-9x+28+6=24\)
\(-x+34=24\)
\(-x=-10\)
\(x=10\)
Vậy \(x=10\)
\(3\left(x-2\right)+2x=10\)
\(3x-6+2x=10\)
\(3x+2x-6=10\)
\(5x=16\)
\(x=\frac{16}{5}\)
Vậy \(x=\frac{16}{5}\)
2(x-1)+3(3x-2)=x-4
=>2x-2=9x-6-x+4=0
=>10x-4=0
=>x=\(\frac{2}{5}\)

c) 135 - | 9 - x | = 3
| 9 - x | = 135 - 3
| 9 - x | = 132
=> 9 - x = 132
x = 9 - 132
x = 123
hoặc 9 - x = - 132
x = 9 - ( - 132 )
x = 9 + 132
x = 141
Vậy ..................
c) 135 - | 9 - x | = 35
\(2\left(x-1\right)+3\left(x-2\right)=x-4\)
\(2x-2+3x-6=x-4\)
\(5x-8=x-4\)
\(5x-8-x+4=0\)
\(4x-4=0\)
\(4x=4\)
\(x=1\)
\(3\left(4-x\right)-2\left(x-1\right)=x+20\)
\(12-x-2x+2=x+20\)
\(14-3x=x+20\)
\(14-3x-x-20=0\)
\(-6-4x=0\)
\(4x=-6\)
\(x=-\frac{3}{2}\)
\(\left|2x+3\right|=5\)
\(\Rightarrow\orbr{\begin{cases}2x=2\\2x=-8\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=-4\end{cases}}}\)

Mọi người giúp em với ạ
Bài tập này em chưa hiểu nên hỏi
Mong mọi người giúp:((
1) x/5 = -12/10
\(x.10=5.\left(-12\right)\)
\(10x=\left(-60\right)\)
\(x=-6\)
2) (x + 1)/2 = x/3
\(3\left(x+1\right)=2x\)
\(3x+3=2x\)
\(3x-2x=3\)
\(x=3\)
3) (1 - 2x)/9 = 1/ (1- 2x)
\(\left(1-2x\right).\left(1-2x\right)=9.1\)
\(\left(1-2x\right)^2=\left(\pm3\right)^2\)
\(\Rightarrow\orbr{\begin{cases}1-2x=3\\1-2x=-3\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=-2\\2x=4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x=2\end{cases}}}\)
4) 4/5 = 12/x = y/20 = 8.(y -x)/z
\(\frac{4}{5}=\frac{12}{x}\Leftrightarrow4x=60\Leftrightarrow x=15\)
\(\frac{4}{5}=\frac{y}{20}\Leftrightarrow y5=80\Leftrightarrow y=16\)
\(\frac{4}{5}=\frac{8.\left(y-x\right)}{z}\Leftrightarrow\frac{4}{5}=\frac{8}{z}\Leftrightarrow4z=40\Leftrightarrow z=10\)
Vậy \(x=15;y=16;z=10\)
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a) \(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(=>2x+\frac{3}{5}=\frac{3}{5}\)
\(2x=\frac{3}{5}-\frac{3}{5}\)
\(2x=0\)
\(x=0:2\)
\(x=0\)
b) \(\left(3x-1\right).\left(-\frac{1}{2x}+5\right)=0\)
=> \(\left(3x-1\right)=0\)hoặc \(\left(-\frac{1}{2x}+5\right)=0\)hoặc \(\left(3x-1\right)\)và\(\left(-\frac{1}{2x}+5\right)\)cùng bằng 0.
\(\orbr{\begin{cases}3x-1=0\\-\frac{1}{2x}+5=0\end{cases}}=>\orbr{\begin{cases}3x=1\\-\frac{1}{2x}=-5\end{cases}}=>\orbr{\begin{cases}x\in\varnothing\\2x=\frac{1}{5}\end{cases}}=>x=\frac{1}{5}:2=>x=\frac{1}{10}\)

a. 4(2x+7)-3(3x-2)=24
=> 8x+28-9x+6=24
=> -x+34=24
=> -x=-10=> x=10
b. 2(x-1)+3(x-2)=x-4
=> 2x-2+3x-6=x-4
=> 5x-8=x-4
=> 5x-x=-4+8
=> 4x=4=> x=1
c. [124-(20-4x)]:30+7=11
=> (124-20+4x):30=4
=> (104+4x):30=4
=> 104+4x=120
=> 4x=96 => x=24

(2x - 7) + 17 = 6
=> 2x - 7 = 6 - 17
=> 2x - 7 = -11
=> 2x = -11 + 7
=> 2x = -4
=> x = -4 : 2
=> x = -2
+) 12 -2(3 - 3x)= -2
=> 2(3 - 3x) = 12 + 2
=> 2(3 - 3x) = 14
=> 3 - 3x = 14 : 2
=> 3 - 3x = 7
=> 3x = 3 - 7
=> 3x = -4
=> x = -4/3
\(\left(x+1\right)\left(x-3\right)=0\)
=> \(\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)
Vậy...
\(\dfrac{1}{2}x-x+20\%x=2\dfrac{1}{4}\\ =>\dfrac{1}{2}x-x+\dfrac{1}{5}x=2+\dfrac{1}{4}\\ =>x\cdot\left(\dfrac{1}{2}-1+\dfrac{1}{5}\right)=\dfrac{9}{4}\\ =>x\cdot\left(\dfrac{5}{10}-\dfrac{10}{10}+\dfrac{2}{10}\right)=\dfrac{9}{4}\\ =>x\cdot\dfrac{-3}{10}=\dfrac{9}{4}\\ =>x=\dfrac{9}{4}:\dfrac{-3}{10}\\ =>x=\dfrac{9}{4}\cdot\dfrac{-10}{3}\\ =>x=\dfrac{-15}{2}\)
\(\dfrac{1}{2}x-x+20\%x=2\dfrac{1}{4}\)
\(\Rightarrow\left(\dfrac{1}{2}-1+\dfrac{1}{5}\right)x=\dfrac{9}{4}\)
\(\Rightarrow-\dfrac{3}{10}x=\dfrac{9}{4}\)
\(\Rightarrow x=\dfrac{9}{4}:\dfrac{-3}{10}\)
\(\Rightarrow x=\dfrac{-15}{2}\)