\((x+y)^3-(x-y)^3\)

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

\(=6x^2y+2y^3\)

Cách khác:

Ta có: \(\left(x+y\right)^3-\left(x-y\right)^3\)

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

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

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

\(=6x^2y+2y^3\)

\(1,P=\left(x+y+x-y\right)\left(x+y-x+y\right)+2\left(x^2-y^2\right)-4y^2\\ P=4xy+2x^2-6y^2\)

Bài 1: 

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

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

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

\(=2x^2+4xy-7y^2\)

(x + y)³ - 3xy(x + y)

= x³ + 3x²y + 3xy² + y³ - 3x²y - 3xy²

= x³ + y³

1. y(y+1)-5y-5                    2. 4x³=x

=y(y+1)-(5y+5)                    <=>4x³-x=0

=y(y+1)-5(y+1)                    <=>x(4x²-1)=0

=(y+1)(y-5)                          <=>x(4x²-1)=0

                                            <=>\(\orbr{\begin{cases}x=0\\4x^2-1=0\end{cases}}\)=\(\orbr{\begin{cases}x=0\\4x^2=1\end{cases}}\)=\(\orbr{\begin{cases}x=0\\x^2=\frac{1}{4}\end{cases}}\)=\(\orbr{\begin{cases}x=0\\x=+_-\frac{1}{2}\end{cases}}\)

3. M= (x+3)² -(4x+1)-x(2x+1)

   M= (x²+6x+9)-4x-1-2x²-x

   M=x²+6x+9-4x-1-2x²-x

   M= -x²+x+8

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

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

c: \(P=4\left(x-3\right)-3\left|x+3\right|\)

Trường hợp 1: x>=-3

\(P=4x-12-3x-9=x-21\)

Trường hợp 2: x<-3

P=4x-12+3x+9=7x-3

= \(\dfrac{1}{9}\cdot x^2\cdot y^3\cdot z\cdot27\cdot y\cdot z^7=3\cdot x^2\cdot y^4\cdot z^8\)

Ta có: \(-\dfrac{1}{9}x^2y^3z\cdot\left(-27yz^7\right)\)

\(=\left[\left(-\dfrac{1}{9}\right)\cdot\left(-27\right)\right]\cdot x^2\cdot\left(y^3\cdot y\right)\cdot\left(z\cdot z^7\right)\)

\(=3x^2y^4z^8\)

a,hđt số 3 = \(\left(a^2+2a\right)^2-9\) 

b,hđt số 3=\(\left[x-\left(y-6\right)\right]\left[x+\left(y-6\right)\right]\)(đổi dấu làm ngoặc khi trước nó là dấu trừ)=\(x^2-\left(y-6\right)^2\)

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

\(=\left(a^2+2a\right)^2+3.\left(-3\right)\)

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

b) \(\left(x-y+6\right)\left(x+y-6\right)\)

\(=\left[x-\left(y-6\right)\right]\left[x+\left(y-6\right)\right]\)

\(=x^2-\left(y-6\right)^2\)