1) \(x^3+y^3+6xy=8\)
\(\Leftrightarrow x^3+y^3+3x^2y+3xy^2+6xy-3x^2y-3xy^2-8=0\)
\(\Leftrightarrow\left(x+y\right)^3-8-3xy\left(x+y-2\right)=0\)
\(\Leftrightarrow\left(x+y-2\right)\left[\left(x+y\right)^2+2\left(x+y\right)+4\right]-3xy\left(x+y-2\right)=0\)
\(\Leftrightarrow\left(x+y-2\right)\left(x^2+y^2-xy+2x+2y+4\right)=0\)
Dễ dàng chứng minh \(\left(x^2+y^2-xy+2x+2y+4\right)>0\forall x;y\)
\(\Rightarrow x+y-2=0\)
\(\Leftrightarrow x+y=2\)
2) \(A=\frac{1}{x^2+y^2}+\frac{1}{xy}\)
\(A=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}\)
Áp dụng bất đẳng thức Cô-si :
\(A\ge\frac{4}{x^2+2xy+y^2}+\frac{1}{2\cdot\frac{\left(x+y\right)^2}{4}}=\frac{4}{\left(x+y\right)^2}+\frac{1}{2\cdot\frac{1}{4}}=\frac{4}{1}+2=6\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=1\)
