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a) \(\left(x+y\right)^2+\left(x-y\right)^2+\left(x+y\right)\left(x-y\right)\)
\(=x^2+2xy+y^2+x^2-2xy+y^2+x^2-y^2\)
\(=3x^2+y^2\)
b)\(\left(3x+y\right)^2+\left(3x-y\right)^2-\left(2x+y\right)\left(2x-y\right)\)
\(=9x^2+6xy+y^2+9x^2-6xy+y^2-4x^2+y^2\)
\(=14x^2+3y^2\)
c) \(2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2\)
\(=\left(x-y\right)^2+2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2\)
\(=\left(x-y+x+y\right)^2\)
\(=4x^2\)
d)\(-2\left(x^2-9y^2\right)+\left(x-3y\right)^2+\left(x+3y\right)^2\)
\(=\left(x+3y\right)^2-2\left(x+3y\right)\left(x-3y\right)+\left(x-3y\right)^2\)
\(=\left(x+3y-x+3y\right)^2=9y^2\)


\(P=\frac{x+3y}{3x+y}.\frac{4x-2y}{x-y}-\frac{x+3y}{3x+y}.\frac{x-3y}{x-y}\)
\(=\frac{x+3y}{3x+y}\left(\frac{4x-2y}{x-y}-\frac{x-3y}{x-y}\right)\)
\(=\frac{x+3y}{3x+y}.\frac{3x+y}{x-y}=\frac{x+3y}{x-y}\)

\(-4.\left(x+3y\right)^3+\left(x-3y\right).\left(x+y\right)\left(x-y\right)-\left(2x-y\right)^3\)
\(=-4.\left(x^3+3.x^2.3y+3.x.9y^2+27y^3\right)+\left(x-3y\right).\left(x^2-y^2\right)-\left(8x^3-3.4x^2.y+3.2x.y^2-y^3\right)\)
\(=-4.\left(x^3+9x^2y+27xy^2+27y^3\right)+\left(x-3y\right).\left(x^2-y^2\right)-\left(8x^3-12x^2y+6xy^2-y^3\right)\)
Còn tiếp...........
\(=-4x^3-36x^2y-108xy^2-108y^3+x^3-xy^2-3x^2y+3y^3-8x^3+12x^2y-6xy^2+y^3\)
\(=-11x^3-27x^2y-115xy^2-104y^3\)
đoạn tiếp nè, chúc cậu học tốt nha ^^

a) \(4\left(x-3y\right).\left(x+3y\right)-\left(2x-y\right)^2\)
\(=4x^2-9y^2-\left(4x^2-4xy+y^2\right)\)
\(=-10y^2+4xy\)

\(2x^3y-2xy^3-4xy^2-2xy\)
\(=2xy.\left(x^2-y^2-2y-1\right)\)
\(=2xy.[x^2-\left(y^2+2y+1\right)]\)
\(=2xy.[x^2-\left(y+1\right)^2]\)
\(=2xy.\left(x+y+1\right).\left(x-y-1\right)\)
Vậy chọn đáp án A
\(\left(x+3y\right)^2-\left(x-y\right)^2\\ =\left(x^2+6xy+9y^2\right)-\left(x^2-2xy+y^2\right)\\ =x^2+6xy+9y^2-x^2+2xy-y^2\\ =\left(x^2-x^2\right)+\left(6xy+2xy\right)+\left(9y^2-y^2\right)\\ =8xy+8y^2\)
Giúp tớ với ah