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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 2 + x - y 2
= x 2 + 2xy + y 2 + x 2 – 2xy + y 2
= 2 x 2 + 2 y 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.
\(A=\left(x-y\right)^2+\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)-4\left(y^2-1\right)\)
\(=\left(x-y-x-y\right)^2-4\left(y^2-1\right)\)
\(=\left(-2y\right)^2-4y^2+4=4\)
= 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
\(\left(x-y\right)^2+2\left(x^2-y^2\right)+\left(x+y\right)^2\)
\(=\left(x-y\right)^2+2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2\)
\(=\left[\left(x-y\right)+\left(x+y\right)\right]^2\)
\(=\left(x-y+x+y\right)^2\)
\(=\left(2x\right)^2\)
\(=4x^2\)
\(\left(x-y\right)^2+2\left(x^2-y^2\right)+\left(x+y\right)^2\)
\(=\left(x-y\right)^2+2\cdot\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2\)
\(=\left(x-y+x+y\right)^2=\left(2x\right)^2=4x^2\)
\(2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2=\left(x+y+x-y\right)^2=\left(2x\right)^2=4x^2\)
\(\left(x+y-7\right)^2-2\left(x+y-7\right)\left(y-6\right)+\left(y-6\right)^2\)
\(=\left(x+y-7-y+6\right)^2\)
\(=\left(x-1\right)^2=x^2-2x+1\)
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
(x + y)2 - (x - y)2
= (x2 + 2xy + y2) - (x2 - 2xy + y2)
= x2 + 2xy + y2 - x2 + 2xy - y2
= (x2 - x2) + (y2 - y2) + (2xy + 2xy)
= 4xy
\(\left(x+y\right)^2-\left(x-y\right)^2=\left(x^2+2xy+y^2\right)-\left(x^2-2xy+y^2\right)\)
\(=x^2+2xy+y^2-x^2-2xy-y^2\)
\(=\left(x^2-x^2\right)+\left(y^2-y^2\right)+\left(2xy+2xy\right)\)
\(=4xy\)