\(P=x^4-64x=x\left(x^3-4^3\right)=x\left(x-4\right)\left(x^2+4x+16\right)\)

P= 125x^3-8y^3

  =5^3x^3-2^3y^3

  =(5x)^3-(2y)^3

  =(5x-2y)(25x^2+10xy+4y^2)

P=4x(x-2y)+8y(2y-x)

  =4x(x-2y)-8y(x-2y)

  =(4x-8y)(x-2y)

  =4(x-2y)(x-2y)

  =4(x-2y)^2

(2x+1)^2-(x-1)^2=(2x+1-x+1)(2x+1+x-1)

                       =(x+2)3x

K NHA!

\(x^2-6xy+9y^2=x^2-2.\left[x.\left(3y\right)\right]+\left(3y\right)^2\)

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

1) \(=x^3\left(x+1\right)+\left(x+1\right)=\left(x+1\right)\left(x^3+1\right)=\left(x+1\right)\left(x+1\right)\left(x^2-x+1\right)\)

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

2) \(=\left(x+1\right)^3-\left(2y\right)^3=\left(x+1-2y\right)\left[\left(x+1\right)^2+\left(x+1\right).2y+4y^2\right]\)

\(=\left(x-2y+1\right)\left(x^2+2x+1+2xy+2y+4y^2\right)\)Đến đây bạn phân tích tiếp nha

CHÚC BẠN HỌC TỐT 

T I C K ủng hộ nha

\(125\left(a+2\right)^3-1\)

\(=\left[5\left(a+2\right)\right]^3-1\)

\(=\left(5a+10\right)^3-1\)

\(=\left(5a+10-1\right)\left[\left(5a+10\right)^2-\left(5a+10\right)+1\right]\)

\(=\left(5a+10-1\right)\left[25a^2+100a+100-5a-10+1\right]\)

\(=\left(5a+9\right)\left[25a^2+95a+91\right]\)

b) \(x^6-1=\left(x^3\right)^2-1=\left(x^3-1\right)\left(x^3+1\right)\)

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

0,125(a+2)³-1

= (0,5)³(a+2)³-1

= (0,5)³(a+2)³-1³

k bt làm nx ==

x⁶-1 = (x³)²-1²=(x³-1)(x³+1)

\(0,125.\left(a+2\right)^3-1\)

\(=0,125.\left(a^3+3.a^2.2+3.a.2^2+2^3\right)-1\)

\(=0,125.\left(a^3+6a^2+12a+8\right)-1\)

\(=0,125a^3+0,75a^2+1,5a+1-1\)

\(=0,125a^3+0,75a^2+1,5a\)

\(x^6-1=\left(x^2\right)^3-1=\left(x^2\right)^3-1^3=\left(x^2-1\right)\left(x^4+x^2+1\right)\)

\(x^7+x^5+1\)

\(\left(x^2+x+1\right)\left(x^5-x^4+x^3-x+1\right)\)

k cho mình nha

\(\text{Đa thức bậc 7 = đa thức bậc 2 + đa thức bậc 5 }\)

\(\text{Chia theo sơ đồ Hoocner sẽ nhanh hơn}\)

\(\text{OoO cô bé tinh nghịch OoO làm đúng rồi nhé}\)

Ta có

a,    x²-x-y²-y

=x²-y²-(x+y)

=(x-y)(x+y) - (x+y)

=(x+y)(x-y-1)

b,   x²-2xy+y²-z²

=(x-y)²-z²

=(x-y-z)(x-y+z)

con bai 32, 33 neu ban tra loi duoc minh h them

nhanh nha

\(x^2-4x+4-y^2\)

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

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

(x - 2)² - y² = (x - 2 - y)(x - 2 + y)

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