Câu 21: 

\(x^2+4x+4=\left(x+2\right)^2\)

Câu 22:

\(x^2-8x+16=\left(x-4\right)^2\)

Ta có: \(81+\left(9-x^2\right)^2\)

Tương tự: \(81+-x^4+18x^2+-81\)

\(=\left(-x^4\right)+\left(18x^2\right)+\left(81+-81\right)\)

\(=-x^4+18x^2\)

\(81-\left(9-x^2\right)^2\)

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

\(=\)\(9^2-\left(9-x^2\right)^2\)

\(=\)\(\left(9-9+x\right)\left(9+9-x\right)\)

\(=\)\(x\left(18-x\right)\)

Chúc bạn học tốt ~ 

81-(9-x^2)^2

=9^2-(3^2-x^2)^2

=9^2-(9- x^2)^2

=(9-9+x).(9+9-x)

​=x.(18-x)

\(9\left(x-3\right)+x\left(x-3\right)=\left(x+9\right)\left(x-3\right)\)

(x-3).(9+x)

2:

-8x^6-12x^4y-6x^2y^2-y^3

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

=-(2x^2+y)^3

3:

=(1/3)^2-(2x-y)^2

=(1/3-2x+y)(1/3+2x-y)

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

                                       \(=\left(2x-\frac{1}{3}\left(y+1\right)\right)\left(2x+\frac{1}{3}\left(y+1\right)\right)\)

                                       \(=\left(2x-\frac{1}{3}y-\frac{1}{3}\right)\left(2x+\frac{1}{3}y+\frac{1}{3}\right)\)

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

x 2  + 6x + 9 =  x 2  + 2.x.3 + 3 2  =  x + 3 2

\(=x^3-27\)

(x - 3) ( x² + 3x + 9 )
= ( x - 3 ).( x² + x.3 + 3² )
= x³ - 3³
= x³ - 27

Bài làm:

Ta có: \(\frac{9}{4x^2}+\frac{9y^2}{4}-\frac{9y}{2x}\)

\(=\left(\frac{3}{2x}\right)^2-2.\frac{3}{2x}.\frac{3y}{2}+\left(\frac{3y}{2}\right)^2\)

\(=\left(\frac{3}{2x}-\frac{3y}{2}\right)^2\)

dạ mk cảm ơn bạn De Bruyne nha!

\(\dfrac{x^2}{4}-3x+9=\left(\dfrac{x}{2}-3\right)^2\)