Ta có :

\(\left(x+y+z\right)^3-x^3-y^3-z^3\)

\(=\left[\left(x+y+z\right)^3-x^3\right]-\left(y^3+z^3\right)\)

\(=\left(x+y+z-x\right)\left[\left(x+y+z\right)^2+x^2+\left(x+y+z\right)x\right]-\left(y+z\right)\left(y^2+z^2-yz\right)\)

\(=\left(y+z\right)\left[x^2+y^2+z^2+2\left(xy+yz+xz\right)+x^2+x^2+xy+zx\right]\)\(-\left(y+z\right)\left(y^2+z^2-yz\right)\)

\(=\left(y+z\right)\left[3x^2+y^2+z^2+3xy+3zx+2yz-y^2-z^2+yz\right]\)

\(=\left(y+z\right)\left[3x^2+3xy+3zx+3yz\right]\)

\(=\left(y+z\right)3\left[\left(x^2+xy\right)+\left(zx+yz\right)\right]\)

\(=3\left(y+z\right)\left(x+y\right)\left(x+z\right)\)

Vậy ...

CÓ AI CHƠI GAME BANG BANG K NÀO

b) \(\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)=0\)

\(\Rightarrow\)

số đó là

0 nhé bn 

mình chỉ

chăc thế thôi

hihi

a) \(x^7+x^2+1\)

\(=x^7-x+x+x^2+1\)

\(=\left(x^7-x\right)+\left(x^2+x+1\right)\)

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

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

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

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

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

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

b) \(x^7+x^5+1\)

\(=x^7+x^6+x^5-x^6+1\)

\(=\left(x^7+x^6+x^5\right)-\left(x^6-1\right)\)

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

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

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

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

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

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

Ta có

   x³-x+y³-y

=(x³+y³)-(x+y)

=(x+y)(x²-xy+y²)-(x+y)

=(x+y)(x²-xy+y²-1)

dễ ợt

a. x^36 - 64 = (x^18)^2 - 8^2 rồi áp dụng hằng đẳng thức số 3 . câu b tương tự

thì 2 câu trên cho dễ vậy để mọi người chú ý 

chứ quan trọng là 2 câu cuối kia kìa