1) \(3\left(5x-2\right)-9=6\)
\(15x-6=6+9\)
\(15x=15+6\)
\(15x=21\)
\(x=\frac{7}{5}\)
2) \(4\left(3x-2\right)+7\left(3x-2\right)=22\)
\(\left(3x-2\right)\left(4+7\right)=22\)
\(\left(3x-2\right)\cdot11=22\)
\(\left(3x-2\right)=2\)
\(3x=4\)
\(x=\frac{4}{3}\)
3) \(2x^2-6x=0\)
\(2x\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x=0\\x-3=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)
Vậy,............
\(1)3\left(5x-1\right)-9=6\)
\(\Rightarrow3\left(5x-2\right)=15\)
\(\Rightarrow5x-2=5\Rightarrow5x=7\Rightarrow x=\frac{7}{5}\)
\(2)4\left(3x-2\right)+7\left(3x-2\right)=22\)
\(\Rightarrow11\left(3x-2\right)=22\)
\(\Rightarrow33x-22=22\Rightarrow33x=44\Rightarrow x=\frac{4}{3}\)
\(3)2x^2-6x=0\)
\(\Rightarrow2x\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
1)
3(5x−2)−9=6
15x−6=6+9
15x=15+6
15x=21
x=\(\frac{7}{5}\)
2)
4(3x−2)+7(3x−2)=22
(3x−2)(4+7)=22
(3x−2)·11=22
(3x−2)=2
3x=4
x=\(\frac{4}{3}\)
32x2−6x=0
2x(x−3)=0
⇒[2x=0
=>[x−3=0
⇒[x=0;x=3
\(3\left(5x-2\right)-9=6\Leftrightarrow3\left(5x-2\right)=6+9\Leftrightarrow3\left(5x-2\right)=15\)
\(\Leftrightarrow5x-2=15:3\Leftrightarrow5x-2=5\Leftrightarrow5x=5+2\Leftrightarrow5x=7\)
\(\Leftrightarrow x=7:5\Leftrightarrow x=\frac{7}{5}\)
\(4\left(3x-2\right)+7\left(3x-2\right)=22\Leftrightarrow\left(3x-2\right)\left(4+7\right)=22\)
\(\Rightarrow\left(3x-2\right).11=22\Leftrightarrow3x-2=22:11\Leftrightarrow3x-2=2\)
\(\Leftrightarrow3x=2-2\Leftrightarrow3x=0\Leftrightarrow x=0:3\Leftrightarrow x=0\)
Phần còn lại tự làm nha bạn !
