\(a,\) Xét \(-\dfrac{34}{35}+\dfrac{2}{5}-\dfrac{3}{7}+1=\dfrac{-34+14-15+35}{35}=0\)
Thay vào ta có: \(\left(2x-3\right)\left(3x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}2x-3=0\Rightarrow x=\dfrac{3}{2}\\3x+1=0\Rightarrow x=-\dfrac{1}{3}\end{matrix}\right.\)
\(b,\) Ta có: \(x^2-5x+6=0\Rightarrow\left(x^2-2x\right)-\left(3x-6\right)=0\)
\(\Rightarrow x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Rightarrow\left(x-3\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x-3=0\Rightarrow x=3\\x-2=0\Rightarrow x=2\end{matrix}\right.\)
`a)`
Tự ghi lại đề
\(\left(2x-3\right)\left(3x+1\right)=\left(\dfrac{8}{2017}-\dfrac{8}{2018}+27\dfrac{3}{2019}\right).0\)
` (2x - 3)(3x + 1) = 0`
\(=>\left[{}\begin{matrix}2x-3=0\\3x+1=0\end{matrix}\right.=>\left[{}\begin{matrix}2x=3\\3x=-1\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
`b`
`x^2 - 5x = -6`
` => x(x-2)-3(x-2) = 0`
`=> ( x - 3)(x -2) = 0`
\(=>\left[{}\begin{matrix}x-3=0\\x-2=0\end{matrix}\right.=>\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)
