Lời giải:

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

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

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

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

   \(=\left(-2y\right)^2-4y^2+4=4\)

(x + y + z)2 – 2.(x + y + z).(x + y) + (x + y)2

= [(x + y + z) – (x + y)]2 (Áp dụng HĐT (2) với A = x + y + z ; B = x + y)

= z2.

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

\(\left(b\right):\left(x-y-z\right)^2+\left(x+y+z\right)^2\\ =\left[\left(x-y\right)-z\right]^2+\left[\left(x+y\right)+z\right]^2\\ =\left(x-y\right)^2-2z\left(x-y\right)+z^2+\left(x+y\right)^2+2z\left(x+y\right)+z^2\\ =x^2-2xy+y^2-2xz+2yz+z^2+x^2+2xy+y^2+2xz+2yz+z^2\\ =2x^2+2y^2+2z^2+4yz\)

\(\left(c\right):\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\\ =\left[\left(x+y\right)-\left(x-y\right)\right]^2\\ =\left(x+y-x+y\right)^2\\ =\left(2y\right)^2=4y^2\)

\(a,\left(x+y\right)^2-\left(x-y\right)^2=\left(x+y-x+y\right)\left(x+y+x-y\right)=4xy\\ b,\left(x+y\right)^2+\left(x-y\right)^2-2\left(x+y\right)\left(x-y\right)=\left(x+y-x+y\right)^2=4y^2\\ c,\left(x^2-1\right)\left(x^2-x+1\right)\\ =\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)\\ =\left(x-1\right)\left(x^3+1\right)\\ =x^4-x^3+x-1\)

a. (x + y)² - (x - y)²

= (x + y - x + y)(x + y + x - y)

= 2y . 2x

= 4xy

b. (x + y)² + (x - y)² - 2(x + y)(x - y)

= (x² + 2xy + y²) + (x² - 2xy + y²) - 2(x² - y²)

= x² + 2xy + y² + x² - 2xy + y² - 2x² + 2y²

= x² + x² - 2x² + 2xy - 2xy + y² + y² + 2y²

= 4y²

c. (x² - 1)(x² - x + 1)

= x⁴ - x³ + x² - x² + x - 1

= x⁴ - x³ + x - 1

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

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

c: \(\left(x^2-1\right)\cdot\left(x^2-x+1\right)\)

\(=\left(x^3+1\right)\left(x-1\right)\)

\(=x^4-x^3+x-1\)

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

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

gúp mình với

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

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

A = x\(^2\) - y\(^2\)

Tại x = \(\dfrac{-1}{2}\) và y = -2 ta có 

\(\left(\dfrac{-1}{2}\right)^2-\left(-2\right)^2\) \(=\) \(\dfrac{-15}{4}\)

\(A=x\left(x+y\right)-y\left(x+y\right)\)

\(=\left(x+y\right)\left(x-y\right)\)

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

Thay \(x=-\dfrac{1}{2}\) và \(y=-2\) vào biểu thức \(A\), ta có:

\(A=\left(-\dfrac{1}{2}\right)^2-\left(-2\right)^2\)

\(=\dfrac{1}{4}-4\)

\(=-\dfrac{15}{4}\)