1) Ta có: 3x-12=5x(x-4)
\(\Leftrightarrow3x-12-5x\left(x-4\right)=0\)
\(\Leftrightarrow3x-12-5x^2+20x=0\)
\(\Leftrightarrow-5x^2+23x-12=0\)
\(\Leftrightarrow-5x^2+20x+3x-12=0\)
\(\Leftrightarrow\left(-5x^2+20x\right)+\left(3x-12\right)=0\)
\(\Leftrightarrow5x\left(-x+4\right)+3\left(x-4\right)=0\)
\(\Leftrightarrow5x\left(4-x\right)-3\left(4-x\right)=0\)
\(\Leftrightarrow\left(4-x\right)\left(5x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4-x=0\\5x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\5x=3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\frac{3}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{4;\frac{3}{5}\right\}\)
2) Ta có: 3x-15=2x(x-5)
\(\Leftrightarrow3x-15-2x\left(x-5\right)=0\)
\(\Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\3-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\2x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\frac{3}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{5;\frac{3}{2}\right\}\)
3) Ta có: 3x(2x-3)+2(2x-3)=0
\(\Leftrightarrow\left(2x-3\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\3x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=\frac{-2}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{3}{2};-\frac{2}{3}\right\}\)
4) Ta có: (4x-6)(3-3x)=0
\(\Leftrightarrow\left[{}\begin{matrix}4x-6=0\\3-3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=6\\3x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{6}{4}=\frac{3}{2}\\x=1\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{3}{2};1\right\}\)
4) (4x - 6 ) ( 3 - 3x ) = 0
<=> \(\left[{}\begin{matrix}4x-6=0\\3-3x=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}4x=6\\3x=3\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=\frac{3}{2}\\x=1\end{matrix}\right.\)
Bài 1 :
a, Ta có : \(3x-12=5x\left(x-4\right)\)
=> \(3x-12=5x^2-20x\)
=> \(3x-12-5x^2+20x=0\)
=> \(5x^2-23x+12=0\)
=> \(5x^2-20x-3x+12=0\)
=> \(5x\left(x-4\right)-3\left(x-4\right)=0\)
=> \(\left(5x-3\right)\left(x-4\right)=0\)
=> \(\left[{}\begin{matrix}5x-3=0\\x-4=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{3}{5}\\x=4\end{matrix}\right.\)
Vậy phương trình có nghiệm là x = \(\frac{3}{5}\) và x = 4 .
b, Ta có : \(3x-15=2x\left(x-5\right)\)
=> \(3x-15-2x\left(x-5\right)=0\)
=> \(3\left(x-5\right)-2x\left(x-5\right)=0\)
=> \(\left(3-2x\right)\left(x-5\right)=0\)
=> \(\left[{}\begin{matrix}3-2x=0\\x-5=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{3}{2}\\x=5\end{matrix}\right.\)
Vậy phương trình có nghiệm là x = \(\frac{3}{2}\) và x = 5 .
c, Ta có : \(3x\left(2x-3\right)+2\left(2x-3\right)=0\)
=> \(\left(3x+2\right)\left(2x-3\right)=0\)
=> \(\left[{}\begin{matrix}3x+2=0\\2x-3=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}3x=-2\\2x=3\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=-\frac{2}{3}\\x=\frac{3}{2}\end{matrix}\right.\)
Vậy phương trình có nghiệm là x = \(-\frac{2}{3}\) và x = \(\frac{3}{2}\) .
d, Ta có : \(\left(4x-6\right)\left(3-3x\right)=0\)
=> \(\left[{}\begin{matrix}4x-6=0\\3-3x=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}4x=6\\-3x=-3\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{6}{4}\\x=1\end{matrix}\right.\)
Vậy phương trình có nghiệm là x = 1 và x = \(\frac{6}{4}\) .
3) 3x(2x - 3 ) + 2(2x - 3 ) =0
<=> (2x - 3) (3x + 2 ) =0
<=> \(\left[{}\begin{matrix}2x-3=0\\3x+2=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}2x=3\\3x=-2\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=\frac{3}{2}\\x=\frac{-2}{3}\end{matrix}\right.\)
Giải:
1) 3x - 12 = 5x(x - 4) ⇔ 3(x - 4) = 5x(x - 4) ⇔ 5x = 3 ⇔ x = \(\frac{3}{5}\)
Vậy: Ngiệm của phương trình là x = \(\frac{3}{5}\)
2) 3x - 15 = 2x(x - 5) ⇔ 3(x - 5) = 2x(x - 5) ⇔ 2x = 3 ⇔ x = \(\frac{3}{2}\)
Vậy: Ngiệm của phương trình là x = \(\frac{3}{2}\)
3) 3x(2x - 3) + 2(2x - 3) = 0 ⇔ 3x(2x - 3) = -2(2x - 3) ⇔ 3x = -2 ⇔ x = \(\frac{-2}{3}\)
Vậy: Ngiệm của phương trình là x = \(\frac{-2}{3}\)
4) (4x - 6)(3 - 3x) = 0 ⇔ \(\left[{}\begin{matrix}4x-6=0\\3-3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=1\end{matrix}\right.\)
Vậy: Ngiệm của phương trình là x = \(\left[{}\begin{matrix}x=\frac{3}{2}\\x=1\end{matrix}\right.\)
