Lời giải:

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

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

(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.

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

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

   \(=\left(-2y\right)^2-4y^2+4=4\)

x + y 2 + x - y 2

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

= 2 x 2  + 2 y 2

= 2(x^2-y^2) + x^2 + 2xy + y^2+x^2-2xy+y^2 
= 2x^2 - 2y^2 + x^2 + 2xy + y^2 + x^2 - 2xy + y^2 
= 4x^2

Theo mình là :

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

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

                                        = x² - y² + x² +y² + (x-y)²

Hãy nhìn kĩ đây là hằng đẳng thức đó
   (x + y)² + 2(x - y)(x + y) + (x - y)²
= (x + y + x - y)²
= (2x)²
= 4x²
 

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

\(\left(x+y-7\right)^2-2\left(x+y-7\right)\left(y-6\right)+\left(y-6\right)^2\)

\(=\left(x+y-7-y+6\right)^2\)

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

2(x – y)(x + y) +  x + y 2  +  x - y 2

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

(áp dụng hằng đẳng thức thứ 1với A = x+ y, B = x- y)

= x + y + x - y 2 = 2 x 2 = 4 x 2

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

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

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

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

\(=\left(2x\right)^2\)

\(=4x^2\)

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

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

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