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Ta có: \(A=\left(x-y-1\right)^3-\left(x-y+1\right)^3+6\left(x-y\right)^2\)
\(=\left(x-y-1-x+y-1\right)\left[\left(x-y-1\right)^2+\left(x-y-1\right)\left(x-y+1\right)+\left(x-y+1\right)^2\right]+6\left(x-y\right)^2\)
\(=-2\cdot\left[3\left(x-y\right)^2+1\right]+6\left(x-y\right)^2\)
\(=-6\left(x-y\right)^2+6\left(x-y\right)^2-2\)
=-2
Bài 1:
a.\(\left(x+y\right)^2-\left(x-y\right)^2=\left(x+y-x+y\right)\left(x+y+x-y\right)=2\left(x+y\right)\)
b.\(2\left(x+y\right)\left(x-y\right)+\left(x+y\right)^2+\left(x-y\right)^2=\left(x+y+x-y\right)^2=4x^2\)
a) \(27\left(1-x\right)\left(x^2+x+1\right)+81x\left(x-1\right)\)
\(=27\left(1-x^3\right)+81\left(x^2-x\right)\)
\(=27-27x^3+81x^2-81x\)
b) \(y\left[x^2+x\left(x-y\right)+\left(x-y\right)^2\right]+\left(x-y\right)^3\)
\(=y\left[x^2+x^2-xy+x^2-2xy+y^2\right]+x^3-3x^2y+3xy^2-y^3\)
\(=y\left(3x^2-3xy+y^2\right)+x^3-3x^2y+3xy^2-y^3\)
\(=3x^2y-3xy^2+y^3+x^3-3x^2y+3xy^2-y^3=x^3\)