\(a^2+2ab+b^2-ac-bc=\left(a+b\right)^2-c\left(a+b\right)=\left(a+b\right)\left(a+b-c\right)\)
Giải:
\(a^2+2ab+b^2-ac-bc=\left(a+b\right)^2-c\left(a+b\right)=\left(a+b\right)\left(a+b-c\right)\)
\(a^2+2ab+b^2-ac-bc=\left(a+b\right)^2-c\times\left(a+b\right)=\left(a+b\right)\times\left(a+b-c\right)\)
a2 + 2ab + b2 - ac - bc
= ( a2 + 2ab + b2 ) - ac - bc
= (a + b ) 2 - c ( a + b )
= ( a+ b ) ( a + b - c )