Câu 21: 

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

Câu 22:

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

    4ⁿ: 2ⁿ⁺¹

= (2²)ⁿ: 2ⁿ⁺¹

= 2²ⁿ:2ⁿ⁺¹

= 2^2n-n-1

= 2^n-1

=(2.2)^n:2^n+1

=2^n.2^n:2^n+1

=2^n+n-n-1

=2^n-1

\(\dfrac{x^2+y}{y^2+x}\)-> (x*x+y)/(y*y+x)

\(x^2+\dfrac{y}{y^2}+x\) -> x*x+y/y*y+x 

x*x và y*y có thể thay thế bằng sqr(x) và sqr(y)

\(1,\\ a,=\left(x+2\right)\left(x^2-2x+4\right)\\ b,=\left(x-4\right)\left(x^2+8x+16\right)\\ c,=\left(3x+1\right)\left(9x^2-3x+1\right)\\ d,=\left(4m-3\right)\left(16m^2+12m+9\right)\\ 2,\\ a,=x^3+125\\ b,=1-x^3\\ c,=y^3+27t^3\)

a)
\(=\left(x+2\right)\left(x^2-2x+4\right)\)
b)
\(=\left(x-4\right)\left(x^2+4x+16\right)\)
c)=\(\left(3x+1\right)\left(9x^2-3x+1\right)\)
d)
=\(\left(4m-3\right)\left(16m^2+12m+9\right)\)

2)
a)
\(=x^3+125\)
\(\)b)\(=1-x^3\)
c)
=\(y^3+27t^3\)
 

\(\frac{9.3^3.1}{81.3^2}=\frac{3.3^3}{3^4.3^2}=\frac{3^4}{3^6}=3^{4-6}=3^{-2}\) . 3^-2 với dạng aⁿ

\(\left(5xy-2\right)\left(5xy+2\right)=\left(5xy\right)^2-2^2=25x^2y^2-4\)

1.24 x 1.24 - 0.24 x 0.24

Đúng nhé , tk nhé , mơn nhìu !!!
~ HOK TỐT ~

Ta có : 

\(1,24^2-0,24^2=\left(1,24+0,24\right).\left(1,24-0,24\right)\)

~ 

`9x^2+4y^2-12xy+6x-4y+1`

`=(3x)^2-2.3x.2y+(2y)^2+2(3x-2y)+1`

`=(3x-2y)^2+2(3x-2y)+1`

`=(3x-2y+1)^2`

9x²+4y²-12xy+6x-4y+1=(3x-2y+1)²