a) x(x-3)+(x-3)=0
<=>(x-3)(x+1)=0
<=>x-3=0 hoặc x+1=0
<=>x=3 hoặc x=-1
b) 7x(x-5)-x+5=0
<=>(7x-1)(x-5)=0
<=>7x-1=0 hoặc x-5=0
<=>x=1/7 hoặc x=5
a: Ta có: \(x\left(x-3\right)+\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
b: Ta có: \(7x\left(x-5\right)-x+5=0\)
\(\Leftrightarrow\left(x-5\right)\left(7x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{1}{7}\end{matrix}\right.\)