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: \(C=x^3+y^3+3xy\left(x^2+y^2\right)-6x^2.y^2\left(x+y\right)\)

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

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

\(=x^3+y^3+3xy.1-6x^2y^2+6x^2y^3\)

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

a) \(\left(x-2y\right)^2+\left(x+2y\right)^2=x^2-4xy+4y^2+x^2+4xy+4y^2=2x^2+8y^2\)

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

\(=2x^2-2y^2+2x^2+2y^2=4x^2\)

\(a,\left(x-2y\right)^2+\left(x+2y\right)^2\)
\(=\left(x^2-4xy+4y^2\right) +\left(x^2+4xy+4y^2\right)\)
\(=2x^2+8y^2\)
\(b,2\left(x-y\right).\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2\)
\(=2\left(x^2-y^2\right)+\left(x^2+2xy+y^2\right)+\left(x^2-2xy+y^2\right)\)
\(=2x^2-2y^2+2x^2+2y^2\)
\(=4x^2\)

a)\(\left(x-2y\right)^2+\left(x+2y\right)^2\)

\(=x^2-4xy+4y^2+x^2+4xy+y^2\)

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

b)\(2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2\)

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

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

2x²+2y²=5xy

<=>2x²-5xy+2y²=0

<=>(2x²-4xy)-(xy-2y²)=0

<=>2x(x-2y)-y(x-2y)=0

<=>(x-2y)(2x-y)=0

<=> x-2y=0 hoặc 2x-y=0

*)Nếu x-2y=0=>x=2y

=>E=\(\frac{x+y}{x-y}=\frac{2y+y}{2y-y}=\frac{3y}{y}=3\)

*)Nếu 2x-y=0=>2x=y

=>E=\(\frac{x+y}{x-y}=\frac{x+2x}{x-2x}=\frac{3x}{-x}=-3\)

Ta có: x>y>0

\(\Rightarrow\hept{\begin{cases}x+y>0\\x-y>0\end{cases}}\)

\(\Rightarrow E=\frac{x+y}{x-y}>0\)

Ta có : E\(=\frac{x+y}{x-y}\)

\(\Rightarrow E^2=\frac{\left(x+y\right)^2}{\left(x-y\right)^2}=\frac{x^2+2xy+y^2}{x^2-2xy+y^2}=\frac{2\left(x^2+2xy+y^2\right)}{2\left(x^2-2xy+y^2\right)}=\frac{2x^2+4xy+2y^2}{2x^2-4xy+2y^2}\)\(=\frac{5xy+4xy}{5xy-4xy}=\frac{9xy}{xy}=9\)

\(\Rightarrow E=\sqrt{9}\)( do E>0)

\(\Leftrightarrow E=3\)

\(A=2x^2+4xy-4x+2y^2-10xy+4y+2xy\)

\(A=\left(2x^2-4xy+2y^2\right)-\left(4x-4y\right)=2\left(x^2-2xy+y^2\right)-4\left(x-y\right)\)

\(A=2\left(x-y\right)^2-4\left(x-y\right)=2.3^2-4.3=6\)

\(\left(2x+y\right)^2=4x^2+4xy+y^2\)

\(\left(x-\frac{y}{2}\right)^2=x^2-xy+\frac{y^2}{4}\)

\(\left(x^2+\frac{y}{2}\right)\left(x^2-\frac{y}{2}\right)=x^4-\frac{y^2}{4}\)

\(\left(x-2y\right)^2\left(x+2y\right)^2=\left(x^2-4y^2\right)^2\)

\(=x^4-8x^2y^2+16y^4\)

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

\(\left(x-2y\right)^2=x^2-4xy+4y^2\)

\(\left(xy^2+1\right)\left(xy^2-1\right)=x^2y^4-1\)

\(\left(x+y\right)^2-4\left(x-y\right)+4=x^2+2xy+y^2-4x+4y+4\)

\(\left(2x+y\right)^2=4x^2+4xy+y^2\)

\(\left(x-\frac{y}{2}\right)^2=x^2-xy+\frac{y^2}{4}\)

\(\left(x^2+\frac{y}{2}\right)\left(x^2-\frac{y}{2}\right)=x^4-\frac{x^2y}{2}+\frac{x^2y}{2}-\frac{y^2}{4}=x^4-\frac{y^2}{4}\)

\(\left(x-2y\right)^2\left(x+2y\right)^2=x^4-8x^2y^2+16y^4\)

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

\(\left(x-2y\right)^2=x^2-4xy+4y^2\)

\(\left(xy^2+1\right)\left(xy^2-1\right)=x^2y^4-xy^2+xy^2-1=x^2y^4-1\)

\(\left(x+y\right)^2-4\left(x-y\right)+4=x^2+2xy+y^2-4x+4y+4\)

ukm, theo giả thiết , ta có y/5=z/5 => y=z sau đó bn tự hỉu nhé ^_^"