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Câu 1: Ta có: A = \(x^3+y^3+3xy=x^3+y^3+3xy\times1=x^3+y^3+3xy\left(x+y\right)\)
\(=\left(x+y\right)^3=1^3=1\)
Câu 2: Ta có: \(B=x^3-y^3-3xy=\left(x-y\right)\left(x^2+xy+y^2\right)-3xy\)
\(=x^2+xy+y^2-3xy=x^2-2xy+y^2=\left(x-y\right)^2=1^2=1\)
Câu 3: Ta có: \(C=x^3+y^3+3xy\left(x^2+y^2\right)-6x^2.y^2\left(x+y\right)\)
\(=x^3+y^3+3xy\left(x^2+2xy+y^2-2xy\right)+6x^2y^2\)
\(=x^3+y^3+3xy\left(x+y\right)^2-3xy.2xy+6x^2y^2\)
\(=x^3+y^3+3xy.1-6x^2y^2+6x^2y^3\)
\(=x^3+y^3+3xy\left(x+y\right)=\left(x+y\right)^3=1^3=1\)
\(x^2+y^2+z^2=xy+yz+zx\)
\(\Leftrightarrow\)\(x^2+y^2+z^2-xy-yz-zx=0\)
\(\Leftrightarrow\)\(2x^2+2y^2+2z^2-2xy-2yz-2zx=0\)
\(\Leftrightarrow\)\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)
\(\Leftrightarrow\)\(\hept{\begin{cases}x-y=0\\y-z=0\\z-x=0\end{cases}}\)\(\Leftrightarrow\)\(x=y=z\)
\(A=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
a, \(\frac{x^2}{x+1}+\frac{2x}{x^2-1}+\frac{1}{x+1}+1\)
\(=\frac{x^2\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\frac{2x}{\left(x-1\right)\left(x+1\right)}+\frac{x-1}{\left(x+1\right)\left(x-1\right)}+\frac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=\frac{x^3-x^2-2x+x-1-x^2-1}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^3-2x^2-x-2}{\left(x-1\right)\left(x+1\right)}\)
a: \(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{x+y}{xy}=\dfrac{5}{-2}=-\dfrac{5}{2}\)
b: \(x^2+y^2=\left(x+y\right)^2-2xy=25-2\cdot\left(-2\right)=29\)
c: \(\dfrac{1}{x^2}+\dfrac{1}{y^2}=\dfrac{x^2+y^2}{\left(xy\right)^2}=\dfrac{29}{4}\)
d: \(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=5^3-3\cdot\left(-2\right)\cdot5=125+6\cdot5=155\)