Question

phân tích đa thức sau hành nhân tử

a) (x+3)(5x-1)=5(x+1)(x-2)

b) (2x-3)(x+1)=2x(x-1)

 

Answer 1

a) \(\left(x+3\right)\left(5x-1\right)=\left(5x+1\right)\left(x-2\right)\)

\(\left(x+3\right)\left(5x-1\right)-\left(5x+1\right)\left(x-2\right)=0\)

\(\left(x+3\right)\left(5x-1\right)+\left(5x-1\right)\left(x-2\right)=0\)

\(\left(x+3+x-2\right)\left(5x-1\right)=0\)

\(\left(2x-1\right)\left(5x-1\right)=0\)

Xảy ra 2 trường hợp:

TH1:2x-1=0⇒x=\(\dfrac{1}{2}\)

TH2:5x-1=0⇒x=\(\dfrac{1}{5}\)

Answer 2

b: Ta có: \(\left(2x-3\right)\left(x+1\right)=2x\left(x-1\right)\)

\(\Leftrightarrow2x^2+2x-3x-3-2x^2+2x=0\)

\(\Leftrightarrow x=3\)