\(1,\\ a,=x^2+2xy+y^2\\ b,=x^2-4xy+4y^2\\ c,=x^2y^4-1\\ d,=\left[\left(x-y\right)\left(x+y\right)\right]^2=\left(x^2-y^2\right)^2=x^4-2x^2y^2+y^4\\ 2,\\ a,=\left(x+2\right)^2\\ b,=\left(3x-2\right)^2\\ c,=\left(\dfrac{x}{2}+1\right)^2\\ d,=\left(x+y-2\right)^2\)

Bài 1 em dùng HĐT nha

Bài 2:

a. x² + 4x + 4

= x² + 2.2.x + 2²

= (x + 2)²

b. 9x² - 12x + 4

= (3x)² - 3x.2.2 + 2²

= (3x - 2)²

c. \(\dfrac{x^2}{4}+x+1\)

= \(\left(\dfrac{x}{2}\right)^2+2.\dfrac{x}{2}.1+1^2\)

= \(\left(\dfrac{x}{2}+1\right)^2\) 

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

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

\(ac-ad+bc-bd\)

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

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

\(ax+by+bx+ay\)

\(=\left(ax+ay\right)+\left(by+bx\right)\)

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

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

ab+ac-do+dc

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

a(x+y)+b(y+x)

Mk làm đúng đó, k mk nha

ab + ac = a(b + c)

ab - ac + ad = a(b - c + d)

ax - bx - cx + dx

=x(a - b - c + d)

chuẩn men

Ta có: \(\left(a^2+b^2\right)\left(c^2+d^2\right)\)

\(=a^2c^2+a^2d^2+b^2c^2+b^2d^2\)

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

\(=\left(ac+bd\right)^2+\left(ad-bc\right)^2\)

=> đpcm

ĐẶT NHÂN TỬ CHUNG NHA!

1) ab - ac + ad = a( b- c +d ) 

2) ax - bx - cx + dx = x( a-b-c+d)

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

4) ac - ad + bc - bd = a( c-d)  + b( c-d) = (a+b)(c-d)

5) ax + by + bx + ay= a( x+y) + b( x+y) = (a+b)(x+y)

a) ab+ac=a(b+c)

b) ab-ac+ad=a(b-c+d)

c) ax-bx-cx+dx = x(a-b-c+d)

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

e) ac-ad+bc-bd = a(c-d)+b(c-d)= (c-d)(a+b)

g) ax+by+bx+ay= x(a+b)+y(a+b)=(x+y)(a+b)

a,a.(b+c)

b,a.(b-c+d)

c,x.(a-b-c+d)

d,(a-d).(b+c)

...............

a,ab+ac=a(b+c)

b,ab-ac+ad=a(b-c+d)

c,ax-bx-cx+dx=(a-b-c+d)x

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

=ab+ac-bd+cd

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

e,ac-ad+bc-bd

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

f,ax+by+bx+ay

=a(x+y)+b(y+x)

a) \(ab+ac=a\left(b+c\right)\)

b) \(ab-ac+ad=a\left(b-c+d\right)\)

c) \(ax-bx-cx+dx=x\left(a-b-c+d\right)\)

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

e) \(ac-ad+bc-bd=a\left(c-d\right)+b\left(c-d\right)=\left(c-d\right)\left(a+b\right)\)

f) \(ax+by+bx+ay=ax+bx+by+ay=x\left(a+b\right)+y\left(a+b\right)=\left(a+b\right)\left(x+y\right)\)