(x+y-2xy).(x+y+2xy)
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a.
(x^2 + y^2 - 2xy) + (x^2 + y^2 + 2xy)
= x^2 + y^2 - 2xy + x^2 + y^2 + 2xy
= (x^2 + x^2) + (y^2 + y^2) + (2xy - 2xy)
= 2x^2 + 2y^2
b.
(x^2 + y^2 - 2xy) - (x^2 + y^2 + 2xy)
= x^2 + y^2 - 2xy - x^2 - y^2 - 2xy
= (x^2 - x^2) + (y^2 - y^2) - (2xy + 2xy)
= -4xy
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\(\dfrac{x}{x^2+2xy+y^2}+\dfrac{2y}{x+y}+\dfrac{y}{x^2+2xy+y^2}\)
\(=\dfrac{x+y}{\left(x+y\right)^2}+\dfrac{2y}{x+y}\)
\(=\dfrac{1}{x+y}+\dfrac{2y}{x+y}=\dfrac{2y+1}{x+y}\)
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a) \(\dfrac{2x^2-2xy}{x^2+x-xy-y}\) \(\left(x\ne y;x\ne-1\right)\)
\(=\dfrac{2x\left(x-y\right)}{x\left(x+1\right)-y\left(x+1\right)}\)
\(=\dfrac{2x\left(x-y\right)}{\left(x-y\right)\left(x+1\right)}\)
\(=\dfrac{2x}{x+1}\)
b) \(\dfrac{x^2+y^2-z^2+2xy}{x^2-y^2+z^2+2xz}\)
\(=\dfrac{\left(x^2+2xy+y^2\right)-z^2}{\left(x^2+2xz+z^2\right)-y^2}\)
\(=\dfrac{\left(x+y\right)^2-z^2}{\left(x+z\right)^2-y^2}\)
\(=\dfrac{\left(x+y+z\right)\left(x+y-z\right)}{\left(x-y+z\right)\left(x+y+z\right)}\)
\(=\dfrac{x+y-z}{x-y+z}\)
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\(\left(x^3+3x^2y+3xy^2+y^3-z^3\right):\left(x+y-z\right)\\ =\left[\left(x+y\right)^3-z^3\right]:\left(x+y-z\right)\\ =\left(x+y-z\right)\left[\left(x+y\right)^2+z\left(x+y\right)+z^2\right]:\left(x+y-z\right)\\ =x^2+2xy+y^2+xz+yz+z^2\)
Vậy chọn A
\(\left(x+y-2xy\right)\left(x+y+2xy\right)\)
\(=\left(x+y\right)^2-4x^2y^2\)
\(=x^2+2xy+y^2-4x^2y^2\)
\(\left(x+y-2xy\right)\left(x+y+2y\right)\)
\(=\left[\left(x+y\right)-2xy\right]\left[\left(x+y\right)+2xy\right]\)
\(=\left(x+y\right)^2-\left(2xy\right)^2\)
\(=\left(x^2+2xy+y^2\right)-4x^2y^2\)