Question

Tính giá trị của bt \(B=\left(\dfrac{x^2-y^2}{xy}-\dfrac{x^3-y^3}{x^2y+xy^2}\right):\dfrac{x-y}{x}\) tại\(x=\dfrac{\sqrt[3]{5}+\sqrt[3]{2}}{\sqrt[3]{5}-\sqrt[3]{2}}\),\(y=\dfrac{\sqrt[3]{2}-\sqrt[3]{5}}{\sqrt[3]{5}+\sqrt[3]{2}}\)

Answer 1

\(B=\dfrac{\left(x^2-y^2\right)\left(x+y\right)-x^3+y^3}{xy\cdot\left(x+y\right)}\cdot\dfrac{x}{x-y}\)

\(=\dfrac{x^3+x^2y-xy^2-y^3-x^3+y^3}{y\left(x+y\right)\left(x-y\right)}\)

\(=\dfrac{xy\left(x-y\right)}{y\left(x+y\right)\left(x-y\right)}=\dfrac{x}{x+y}\)

\(x+y=\dfrac{\sqrt[3]{25}+2\sqrt[3]{10}+\sqrt[3]{4}-\sqrt[3]{25}+2\sqrt[3]{10}-\sqrt[3]{4}}{\sqrt[3]{25}-\sqrt[3]{4}}=\dfrac{4\sqrt[3]{10}}{\sqrt[3]{25}-\sqrt[3]{4}}\)

=>\(D=\dfrac{\sqrt[3]{5}+\sqrt[3]{2}}{\sqrt[3]{5}-\sqrt[3]{2}}:\dfrac{4\sqrt[3]{10}}{\sqrt[3]{25}-\sqrt[3]{4}}\)