\(\text{•}3a-3b+a^2-2ab+b^2=3\left(a-b\right)+\left(a-b\right)^2\\ =\left(a-b\right)\left(3+a-b\right)\\\)
câu 2 sửa đề nhá: \(a^2+2ab+b^2-2a-2b+1\)
\(\text{•}a^2+2ab+b^2-2a-2b+1\\ =\left(a+b\right)^2-2\left(a+b\right).1+1^2\\ =\left(a+b-1\right)^2\)
Phân tích đa thức thành nhân tử
\(3a-3b+a^2-2ab+b^2=\)
\(=\left(3a-3b\right)+\left(a^2-2ab+b^2\right)\)
\(=3\left(a-b\right)+\left(a-b\right)^2\)
\(=3\left(a-b\right)+\left(a-b\right)\left(a-b\right)\)
\(=\left(a-b\right)\left(3+a-b\right)\)
\(3a-3b+a^2-2ab+b^2\)
\(=\left(3a-3b\right)+\left(a^2-2ab+b^2\right)\)
\(=3\left(a-b\right)+\left(a-b\right)^2\)
\(=\left(a-b\right)\left(3+a-b\right)\)
