\(x^3-y^3-6xy\)=\(x^3-3x^2y+3xy^2-y^3-6xy+3x^2y-3xy^2\)=\(\left(x-y\right)^3+3xy\left(-2+x-y\right)\)

Thay x-y=2 vào biểu thức.

=>\(2^3+3xy\left(-2+2\right)\)=\(8+0=8\)

 x³ - y³ - 6xy = \(\left(x-y\right)\left(x^2+xy+y^2\right)-6xy\)

thay x - y = 2 vào 

\(2\left(x^2+xy+y^2\right)-6xy=2x^2+2xy+2y^2-6xy=2x^2-4xy+2y^2=2\left(x^2-2xy+y^2\right)=2\left(x-y\right)^2\)thay x- y = 2

2 . 2^2 = 2 . 4 = 8

P/s: Ko chắc lắm.

\(A=x^3+y^3+6xy-3x-3y+1\)

\(A=\left(x+y\right)\left(x^2-xy+y^2\right)-3\left(x+y\right)+6xy+1\)

\(A=\left(x+y\right)\left(x^2+2xy+y^2-2xy-xy\right)-3\left(x+y\right)+6xy+1\)

\(A=\left(x+y\right)\left[\left(x+y\right)^2-3xy\right]-3\left(x+y\right)+6xy+1\)

\(A=\left(x+y\right)\left[\left(x+y\right)^2-3xy-3\right]+6xy+1\)

Thay x+y=2 vào biểu thức, ta có:

\(A=2\left(2^2-3xy-3\right)+6xy+1\)

\(A=2\left(1-3xy\right)+6xy+1\)

\(A=2-6xy+6xy+1\)

\(A=3\)

\(B=x^2-y^2+4y+1\)

\(B=\left(x-y\right)\left(x+y\right)+4y+1\)

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

\(B=2x-2y+4y+1\)

\(B=2x+2y+1\)

\(B=2\left(x+y\right)+1=2.2+1=5\)

A=x³-y³-3xy(x-y)=(x-y)³=8

\(A=\left(x^3-y^3\right)-6xy\)

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

\(A=2x^2+2xy+2y^2-6xy\)

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

\(A=2.2^2=2.4=8\)

Vì \(x-y=2\)
\(\Leftrightarrow x=2+y\)(1)

Thế (1) vào \(A\)ta có :
 \(A=\left(2+y\right)^3-6\left(2+y\right)y-y^3\)

\(A=8+12y+6y^2+y^3-12y-6y^2-y^3\)

\(A=8\)

Đáp án : 8

\(x^2-6xy+9y^2\)

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

\(=\left[15-3.\left(-3\right)\right]^2\)

\(=\left(15+9\right)^2\)

\(=24^2\)

\(=576\)

\(\dfrac{x^2-9y^2}{x^2-6xy+9y^2}\) tại x = 1 , y = -\(\dfrac{2}{3}\)

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

= \(\dfrac{\left(x-3y\right)\left(x+3y\right)}{\left(x-3y\right)}\)

= (x + 3y)

 Thay x = 1 , y = -\(\dfrac{2}{3}\) vào 

   x + 3y 

= 1 +3 . -\(\dfrac{2}{3}\)

= -1

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