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

\(a,=\left(3+x\right)\left(9-3x+x^2\right)\\ b,=\left(4x+0,1\right)\left(16x^2-0,4x+0,01\right)\\ c,=\left(2-3x\right)\left(4+6x+9x^2\right)\\ d,=\left(\dfrac{x}{5}-\dfrac{y}{3}\right)\left(\dfrac{x^2}{25}+\dfrac{xy}{15}+\dfrac{y^2}{9}\right)\)

a) \(27+x^3=3^3+x^3=\left(3+x\right)\left(9-3x+x^2\right)\)

b) \(64x^3+0,001=\left(4x\right)^3+\left(\dfrac{1}{10}\right)^3=\left(4x+\dfrac{1}{10}\right)\left(16x^2-\dfrac{4x}{10}+\dfrac{1}{100}\right)\)

a/\(27+x^3=\left(3+x\right)\left(9-3x+x^2\right)\)

b/ \(64x^3+0,001=\left(4x+0,1\right)\left(16x^2-0,4x+0,01\right)\)

c/ \(8-27x^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)

d/ \(\dfrac{x^3}{125}-\dfrac{y^3}{27}=\left(\dfrac{x}{5}-\dfrac{y}{3}\right)\left(\dfrac{x^2}{25}+\dfrac{xy}{15}+\dfrac{y^2}{9}\right)\)

2x y 2  +  x 2 y 4 + 1 = x y 2 2 + 2.x y 2 .1 + 1 2  = x y 2 + 1 2

b: \(3^4\cdot3^5:\dfrac{1}{27}==3^9\cdot3^3=3^{12}\)

a,16x²-9

=4²x²-3²

=(4x-3)(4x+3) HĐT thứ 3

b,9a²-25b²

=3²a²-5²b²

=(3a-5b)(3a+5b) HĐT thứ 3

c,81-y⁴

=3².3²-y².y²

=(3²-y²)

=(3-y)(3+y) HĐT thứ 3

d,(2x+y)²-1

=(2x+y-1)(2x+y-1) HĐT thứ 3

e,(x+y+z)²-(x-y-z)²

cái này là HĐT thứ 8 mở rộng bạn lên mạng tìm nha

a) \(16x^2-9=\left(4x\right)^2-3^2=\left(4x-3\right).\left(4x+3\right)\)

b) \(9a^2-25b^4=\left(3a\right)^2-\left(5b^2\right)^2=\left(3a-5b^2\right).\left(3a+5b^2\right)\)

c) \(81-y^4=9^2-\left(y^2\right)^2=\left(9-y^2\right).\left(9+y^2\right)\)

d)\(\left(2x+y\right)^2-1=\left(2x+y\right)^2-1^2=\left(2x+y-1\right).\left(2x+y+1\right)\)

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)

Câu 21: 

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

Câu 22:

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

\(a,=8\left(x^3-125\right)=8\left(x-5\right)\left(x^2+5x+25\right)\\ b,=\left(0,1+4x\right)\left(0,01-0,4x+16x^2\right)\\ c,=\left(x+\dfrac{1}{5}y\right)\left(x^2-\dfrac{1}{5}xy+\dfrac{1}{25}y^2\right)\\ d,=\left(3x-\dfrac{1}{2}y\right)\left(9x^2+\dfrac{3}{2}xy+\dfrac{1}{4}y^2\right)\\ e,=\left(x-1+3\right)\left[\left(x-1\right)^2-3\left(x-1\right)+9\right]\\ =\left(x+2\right)\left(x^2-2x+1-3x+3+9\right)\\ =\left(x+2\right)\left(x^2-5x+13\right)\\ f,=\left(\dfrac{x^2}{2}-y^2\right)\left(\dfrac{x^4}{4}+\dfrac{x^2y^2}{2}+y^4\right)\)