`3x(x-5)+4(5-x)=0`
`<=> 3x(x-5)-4(x-5)=0`
`<=>(x-5)(3x-4)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\Leftrightarrow x=5\\3x-4=0\Leftrightarrow x=\dfrac{4}{3}\end{matrix}\right.\)
Đề yêu cầu gì vậy em?
\(3x\left(x-5\right)+4\left(5-x\right)=0\)
⇔ \(3x\left(x-5\right)-4\left(x-5\right)=0\)
⇔ \(\left(x-5\right)\left(3x-4\right)=0\)
⇒ \(x-5=0\) hoặc \(3x-4=0\)
\(1.x-5=0\Leftrightarrow x=5\)
\(2.3x-4=0\Leftrightarrow x=\dfrac{4}{3}\)
Vậy \(S=\left\{5,\dfrac{4}{3}\right\}\)
\(3x\left(x-5\right)+4\left(5-x\right)=0\)
\(\Leftrightarrow3x\left(x-5\right)-4\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(3x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\3x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\3x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{4}{3}\end{matrix}\right.\)
\(\text{Vậy phương trình có tập nghiệm là }S=\left\{5;\dfrac{4}{3}\right\}\)
