b; 1³ = (\(x-y\))³ = \(x^3\) - 3\(x^2\).y + 3\(xy^2\) - y³ = \(x^3\) - y³ - 3\(xy\)(\(x-y\)) 

    1 = \(x^3\) - y³ - 3\(xy\)

a)  \(x+y=1\)

=>   \(\left(x+y\right)^3=1\)

<=>  \(x^3+y^3+3xy\left(x+y\right)=1\)

<=>  \(x^3+y^3+3xy=1\)

b)  \(x-y=1\)

=>  \(\left(x-y\right)^3=1\)

<=>  \(x^3-y^3-3xy\left(x-y\right)=1\)

<=>  \(x^3-y^3-3xy=1\)

x^3+ y^3+ 3xy

=(x+y)(x^2 -xy + y^2 ) + 3xy
=x^2  -xy + y^2 + 3xy

=x^2 + 2xy + y^2

=(x+y)^2 =1

=> x^3+ y^3+ 3xy=1

còn câu b ai giúp m vs

1³ = (\(x+y\))³ = \(x^3\) + 3\(x^2\)y + 3\(xy^2\) + y³ = \(x^3\)+y³+3\(xy\)(\(x+y\))

1 = \(x^3\)+y³+3\(xy\)

1³ = (\(x-y\))³ = \(x^3\) - 3\(x^2\)y + 3\(xy\) - y³ = \(x^3\) - y³ - 3\(xy\)(\(x-y\))

1 = \(x^3\) - y³ - 3\(xy\)

a)\(A=\left(\frac{x+y}{x-2y}+\frac{3y}{2y-x}-3xy\right).\frac{x+1}{3xy-1}+\frac{x^2}{x+1}\)

\(=\left(\frac{x+y-3y}{x-2y}-3xy\right).\frac{x+1}{3xy-1}+\frac{x^2}{x+1}\)

\(=\left(\frac{x-2y}{x-2y}-3xy\right).\frac{x+1}{3xy-1}+\frac{x^2}{x+1}\)

\(=\left(1-3xy\right).\frac{-x-1}{1-3xy}+\frac{x^2}{x+1}\)

\(=-\left(x+1\right)+\frac{x^2}{x+1}\)`

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

\(=\frac{-x^2-2x-1+x^2}{x+1}\)

\(=\frac{-2x-1}{x+1}\)(1)

b) Thay \(x=-3,y=2014\)vào (1) ta được:

\(A=\frac{-2.\left(-3\right)-1}{-3+1}=\frac{-5}{2}\)

Vậy \(A=\frac{-5}{2}\)với x=-3 và y=2014

   \(A=x^3+3xy+y^3\)

       \(=\left(x+y\right)\left(x^2-xy+y^2\right)+3xy\)

       \(=1.\left(x^2-xy+y^2\right)+3xy\)

       \(=x^2-xy+y^2+3xy\)

       \(=x^2+2xy+y^2\)

       \(=\left(x+y\right)^2\)

       \(=1\)

\(x+y=1\)

\(\Leftrightarrow\)\(\left(x+y\right)^2=1\)

\(\Leftrightarrow\)\(x^2+y^2=1-2xy\)

\(x+y=1\)

\(\Leftrightarrow\)\(\left(x+y\right)^3=1\)

\(\Leftrightarrow\)\(x^3+y^3=1-3xy\)

\(H=1-3xy+3xy\left(1-2xy\right)+6x^2y^2\left(xy+y\right)\)

\(=1-6x^2y^2+6x^2y^2\left(xy+y\right)\)

\(=1-6x^2y^2\left(1-xy-y\right)\)

\(=1-6x^2y^2\left(x+y-xy-y\right)\)

\(=1-6x^2y^2\left(x-xy\right)\)

\(=1-6x^3y^2\left(1-y\right)\)

\(=1-6x^3y^2\left(x+y-y\right)\)

\(=1-6x^4y^2\)

mới ra đc đến đây

Ta có:
\(x^3+3xy-y^3=x^3-y^3+3xy=\left(x-y\right)\left(x^2+xy+y^2\right)+3xy\)

\(=-\left(x^2+xy+y^2\right)+3xy=-x^2-xy+y^2+3xy=-x^2+2xy+y^2=y^2+2xy-x^2\)
\(=-\left(y^2-2xy+x^2\right)=-\left(y-x\right)^2=-\left(x-y\right)^2=-\left(-1\right)^2=-1\)

tick đúng nha

x³+y³=x³+3x²y+3xy²+y²+3xy-3x²y-3xy²

=(x+y)³+3xy.(1-x-y)

=(x+y)³+3xy.[1-(x+y)]

=1³+3xy.(1-1)

=1

1³ - 3xy . (1-1) = 1 

>_< chúc bn học tốt