\(\Leftrightarrow\left(a+b\right)^2+2\left(a+b\right)+1=\left(a+b+1\right)^2\)

1) a^2 + b^2 + 2a - 2b - 2ab = (a^2 - 2ab + b^2) + (2a-2b) = (a-b)^2 + 2(a-b) = (a-b)(a-b+2)

2) 4a^2 - 4b^2 - 4a + 1 = ( 4a^2 - 4a +1) - 4b^2 = (2a-1)^2 - 4b^2 = (2a-1-2b)(2a-1+2b)

3) a^3+6a^2+12a+8= (a^3+8)+(6a^2+12a)= (a+2)(a^2-2a+4)+6a(a+2)=(a+2)(a^2-2a+4+6a)=(a+2)(a^2+4a+4)=(a+2)(a+2)^2=(a+2)^3

a) \(a^2+b^2+2ab+2a+2b+1=\left(a^2+2ab+b^2\right)+\left(2a+2b\right)+1\)

\(=\left(a+b\right)^2+2\left(a+b\right)+1=\left[\left(a+b\right)+1\right]^2=\left(a+b+1\right)^2\)

b) K phân tích dc.

giúp với ạ

a) \(3a-3b+a^2-2ab+b^2\)

\(=3\left(a-b\right)+\left(a-b\right)^2\)

\(=\left(a-b\right)\left(a-b+3\right)\)

a)

3.(a-b) +2.(a-b ) =5 .(a-b )

câu b làm tương tự nha nhóm a^2 -2ab +b^2 vào 1nhoms và làm như câu a

mình làm nhầm 

a) \(\left(a-b\right).\left(a-b+3\right)\)

a² -2ab+2b+b²-2a=(a²-2ab+b²)-(2a-2b)=(a-b)²-2(a-b)=(a-b)(a-b-2)

x^2 - y^2 - 2x + 4y - 3

= (x^2 - 2x + 1) - (y^2 - 4y + 4)

=  (x + 1)^2 - (y + 2)^2

= (x + 1 - y - 2)(x + 1 + y + 2)

= (x - y - 1)(x + y + 3)

a^2 + b^2 - 2a + 2b - 2ab

= (a^2 - 2ab + b^2) - 2(a - b)

= (a - b)^2 - 2(a - b)

= (a - b)(a - b - 2)

a^2+b^2-2a+2b-2ab

=(a^2+b^2-2ab)-(2a-2b)

=(a-b)^2-2(a-b)

=(a-b)(a-b-2)

Ta có : a² + b² - 2a + 2b - 2ab

= ( a² - 2ab + b² ) - ( 2a - 2b )

= ( a - b )² - 2( a - b )

= ( a - b )( a - b - 2 )

a, \(\left(x+1\right)^2-2\left(x+1\right)\left(y-3\right)+\left(y-3\right)^2=\left[\left(x+1\right)-\left(y-3\right)\right]^2\)

\(=\left(x+1-y+3\right)^2=\left(x-y+4\right)^2\)

b, \(a^2+b^2+2a-2b-2ab=\left(a^2-2ab+b^2\right)+\left(2a-2b\right)\)

\(=\left(a-b\right)^2+2\left(a-b\right)=\left(a-b\right)\left[\left(a-b\right)+2\right]=\left(a-b\right)\left(a-b+2\right)\)