(b-a-5)*(b-a)

a²+b²-5b-2ab+5a
= (b²-2ab+a²)-5(b-a)
= (b-a)²-5(b-a)
= (b-a)(b-a-5)

= ( a + b ) mũ 2 - c ( a + b )

= ( a + b ) ( a + b - c )

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

\(2ab^2-a^2b-b^3\)

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

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

-(2ab² - a²b - b³)

= b(-2ab + a² + b²)

= b(a² - 2ab + b²)

= b(a - b)²

a² + 2ab + b² - ac - bc 

= ( a² + 2ab + b² ) - ac - bc 

= ( a + b ) ² - c ( a - b ) 

\(a^2+2ab+b^2-ac-bc=\left(a+b\right)^2-c\left(a+b\right)\)

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

Chúc bạn học tốt.

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

toán lớp 8 nhưg mk mới học lớp 7 thôi!

thông cảm! mik ko thể giúp bn đc

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

\(2ab^2-a^2b-b^3\)

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

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

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

\(25-a^2+2ab-b^2\)

\(=25-\left(a^2-2ab+b^2\right)\)

\(=5^2-\left(a-b\right)^2\)

\(=\left(5-a+b\right)\left(5+a-b\right)\)

\(=25-\left(a-b\right)^2\)

\(=\left(25-a+b\right)\left(25+a-b\right)\)

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

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

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

(a-b)²

a: =x^2+2xy+y^2-4x^2y^2

=(x+y)^2-(2xy)^2

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

b: =49-(a^2-2ab+b^2)

=49-(a-b)^2

=(7-a+b)(7+a-b)

c: =\(a^2-\left(b^2-4bc+4c^2\right)\)

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

d: 

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

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

e: \(=\left(a+b\right)^2+2c\left(a+b\right)+c^2+\left(a+b\right)^2-2c\left(a+b\right)+c^2-4c^2\)

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

=2[(a+b)^2-c^2]

=2(a+b-c)(a+b+c)