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

=-3xy+2x+y+1

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

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

   \(=\left(-2y\right)^2-4y^2+4=4\)

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

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

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

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

\(=-xy\)

Vậy \(A=-xy\)

#\(Toru\)

\(A=\dfrac{\left(x+y\right)^2-z^2}{x+y+z}\)

Đk: \(x\ne y\ne z\)

\(\Rightarrow A=\dfrac{\left(x+y+z\right)\left(x+y-z\right)}{x+y+z}\)

        \(=x+y-z\)

Lời giải:

\(A=\frac{x^3-y^3-z^3-3xyz}{(x+y)^2+(y-z)^2+(x+z)^2}=\frac{(x-y)^3+3xy(x-y)-z^3-3xyz}{x^2+y^2+2xy+y^2-2yz+z^2+z^2+x^2+2xz}\)

\(=\frac{(x-y)^3-z^3+3xy(x-y-z)}{2x^2+2y^2+2z^2+2xy-2yz+2xz}=\frac{(x-y-z)[(x-y)^2+z(x-y)+z^2]+3xy(x-y-z)}{2(x^2+y^2+xy-yz+xz)}\)

\(=\frac{(x-y-z)[(x-y)^2+z(x-y)+z^2+3xy]}{2(x^2+y^2+xy-yz+xz)}=\frac{(x-y-z)(x^2+y^2+z^2+xy-yz+xz)}{2(x^2+y^2+z^2+xy-yz+xz)}=\frac{x-y-z}{2}\)

x + y 2 + x - y 2

= x 2  + 2xy + y 2  +  x 2  – 2xy +  y 2

= 2 x 2  + 2 y 2

Lời giải:

$x(x+y)-y(x+y)+x^2+y^2=(x-y)(x+y)+x^2+y^2$

$=x^2-y^2+x^2+y^2=2x^2$

a, \(A=4xy^2\)

Thay x = 3 ; y = -2 ta đc 

4 . 3 (-2)^2 = 12 . 4 = 48 

a) A = x - y + z + z + y + x - 2y

A = (x + x) + (-y + y) + (z + z) - 2y

A = 2x + 0 + 2z - 2y 

A = 2 .(x + z - y)

b) Thay x = 3 ; y = -1 ; z = 2 vào biểu thức A , ta được :

A = 2 .[3 + 2 - (-1)]

A = 12

Vậy A = 12

Chúc bạn học tốt !