\(a^2c^2-acd+abc-bd\)

\(=ac\left(ac-d\right)+b\left(ac-d\right)\)

\(=\left(ac-d\right)\left(ac+b\right)\)

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

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

c) \(a^2-2a+1=\left(a-1\right)^2=\left(a-1\right)\left(a-1\right)\)

tịnh vượn

\(a^2-b^2=\text{aa}.bb=\left(a-b\right).\left(a+b\right)\)

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

\(a^2-2a+1=\left(a-1\right)^2=\left(a-1\right).\left(a-1\right)\)

a ^2 + b^2 + 2ab 

= a^2 + 2ab + b^2 ( viết xuôi) ( ko cần thiết) 

= (a+b)^2 

= (a+b)(a+b) ( ko cần thiết)

vì lũy thữa cũng coi là 1 tích

a² + b² + 2ab = (a + b) ^{_{(a + b) = (a + b)} 2}

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

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

a) ab+bd-ac-cd

= (ab + bd) - (ac + cd)

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

= (b-c)(a+d)

b) ax+by-ay-bx

= (ax - ay) - (bx - by)

= a(x-y) - b(x-y)

= (a-b)(x-y)

c) x²-xy-xy+y²

= x^2 - 2xy + y^2

= (x-y)^2

3a²c²+bd+3abc+acd=3ac(ac+b)+d(ac+b)=(3ac+d)(ac+b)

\(x^2-6x+9=x^2-2.3x+3^2=\left(x-3\right)^2\)

\(\frac{1}{4}a^2+2ab^2+4b^4=\left(\frac{1}{2}a\right)^2+2.\frac{1}{2}a.2b^2+\left(2b\right)^2=\left(\frac{1}{2}a+2b\right)^2\)

\(25+10x+x^2=5^2+2.5x+x^2=\left(5+x\right)^2\)

\(\frac{1}{9}-\frac{2}{3}y^4+y^8=\left(\frac{1}{3}\right)^2-2.\frac{1}{3}y^4+\left(y^4\right)^2=\left(\frac{1}{3}-y^4\right)^2\)

a,(x-3)^2

b,(1/4x+2b^2)^2

c,(5+x)^2

d,(1/3-y^4)^2

sorry ,phần b fải là (1/2a+2b^2)^2

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