Question

[Ongniel]

Tính \(E=\left(-2016\right).20152015+20162016.2015\)

Answer 1

[ღღ_๖ۣ nhók_lùn ❣_ღღ]

E = ( -2016 ) . 20152015 + 20162016 . 2015

E = ( -2016 ) . 2015 . 10001 + 2016 . 10001 . 2015

E = 2015 . ( ( -2016 ) . 10001 + 2016 . 10001 )

E = 2015 . 0

E = 0

Vậy E = 0

Answer 2

[Hattori Heiji]

$A=\dfrac{x^2}{(x+y)(1-y)}-\dfrac{y^2}{(x+y)(1+x)}-\dfrac{x^2y^2}{(1+x)(1-y)}$

$A=\dfrac{x^2(x+1)-y^2(1-y)-x^2y^2(x+y)}{(x+y)(1+x)(1-y)}$

$A=\dfrac{x^3+x^2-y^2+y^3-x^3y^2-x^2y^3}{(x+y)(1+x)(1-y)}$

$A=\dfrac{(x^3+y^3)+(x^2-y^2)-(x^3y^2+x^2y^3)}{(x+y)(1+x)(1-y)}$

$A=\dfrac{x^3(x^2-xy+y^2)+(x-y)(x+y)-x^2y^2(x+y)}{(x+y)(1+x)(1-y)}$

$A=\dfrac{(x+y)(x^2-xy+y^2+x-y-x^2y^2)}{(x+y)(1+x)(1-y)}$

$A=\dfrac{(x+y)[(x^2+x)-(xy+y)+(y^2-x^2y^2)}{(x+y)(1+x)(1-y)}$

$A=\dfrac{(x+y)[x(x+1)-y(x+1)+y^2(1-x)(1+x)]}{(x+y)(1+x)(1-y)}$

$A=\dfrac{(x+y)(1+x)(x-y+y^2-xy^2}{(x+y)(1+x)(1-y)}$

$A=\dfrac{(x+y)(1+x)[x(1-y)(1+y)-y(1-y)}{(x+y)(1+x)(1-y)}$

$A=\dfrac{(x+y)(1+x)(1-y)(x+xy-y)}{(x+y)(1+x)(1-y)}$

$A=x+xy-y$