Question

cho x,y thỏa mãn \(x^3-x^2+x-5=0\) và \(y^3-2y^2+2y+4=0\)

tinhs S=x+y

Answer 1

\(y^3-2y^2+2y+4=0\Leftrightarrow\left(y-1\right)^3+y^2-y+5=0\)

\(\Leftrightarrow x^3+\left(y-1\right)^3-x^2+x+y^2-y=0\)

\(\Leftrightarrow\left(x+y-1\right)\left(x^2-x\left(y-1\right)+\left(y-1\right)^2\right)-\left(x-y\right)\left(x+y-1\right)=0\)

\(\Leftrightarrow\left(x+y-1\right)\left(x^2-x\left(y-1\right)+\left(y-1\right)^2-x+y\right)=0\)

Mà \(x^2-x\left(y-1\right)+\left(y-1\right)^2-x+y=x^2-xy+y^2-y+1=\left(x-\dfrac{y}{2}\right)^2+\dfrac{3}{4}\left(y-\dfrac{2}{3}\right)^2+\dfrac{2}{3}>0\)

\(\Rightarrow x+y-1=0\Rightarrow x+y=1\)