Question

Cho x-y = 4 ; x² + y² = 12. Tính A = x³ - y³

Answer 1

có `x-y=4`

`<=>x^2 -2xy+y^2 =16`

`<=>12-2xy+16`

`<=>-2xy=4`

`<=>xy=-2`

 

`x^3 -y^3`

`=(x-y)(x^2 +xy+y^2)`

`=4(12-2)`

`=4*10`

`=40`

Answer 2

Có: \(x-y=4\)
\(\Rightarrow\left(x-y\right)^2=16\)

\(\Rightarrow x^2-2xy+y^2=16\)

\(\Rightarrow x^2+y^2-2xy=16\)

\(\Rightarrow12-2xy=16\) \(\Leftrightarrow2xy=-4\Leftrightarrow xy=-2\)

Lại có: \(A=x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)\)

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

\(=4.\left(12-2\right)\) (do \(x^2+y^2=12;xy=-2\))

\(=4.10=40\)

                                                                            Vậy \(A=40\).