Question

cho xy =11 và x^2y+xy^2+x+y= 2010 tình c=x^2+y^2

 

Answer 1

\(x^2y+xy^2+x+y=2010\)

\(\Rightarrow xy\cdot\left(x+y\right)+x+y=2010\)

\(\Rightarrow\left(xy+1\right)\cdot\left(x+y\right)=2010\)

Với : \(xy=11\)

\(\Rightarrow x+y=\dfrac{2010}{12}=\dfrac{335}{2}\)

\(C=x^2+y^2=\left(x+y\right)^2-2xy=\left(\dfrac{335}{2}\right)^2-2\cdot11=\dfrac{112137}{4}\)

Answer 2

Ta có: \(x^2y+xy^2+x+y=2010\)

\(\Leftrightarrow xy\left(x+y\right)+\left(x+y\right)=2010\)

\(\Leftrightarrow\left(x+y\right)\left(xy+1\right)=2010\)

\(\Leftrightarrow x+y=\dfrac{2010}{11+1}=\dfrac{2010}{12}=\dfrac{335}{2}\)

Ta có: \(C=x^2+y^2\)

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

\(=\left(\dfrac{335}{2}\right)^2-2\cdot11\)

\(=\dfrac{112137}{4}\)