Question

cho \(a=\sqrt[3]{3+\sqrt[3]{3}}+\sqrt[3]{3-\sqrt[3]{3}}\)và \(b=2\sqrt[3]{3}\), so sánh a và b

Answer 1

Đặt \(x=\sqrt[3]{3+\sqrt[3]{3}};y=\sqrt[3]{3-\sqrt[3]{3}}\) => \(x^3+y^3=3+\sqrt[3]{3}+3-\sqrt[3]{3}=6\)

Ta có: \(b^3-a^3=\left(2\sqrt[3]{3}\right)^3-\left(x+y\right)^3=24-\left(x+y\right)^3\)

= 4(x³ + y³) - (x³ + y³) - 3xy(x + y)

= 3(x³ + y³) - 3xy(x + y)

= 3(x + y)(x² - xy + y²) - 3xy(x + y)

= 3(x + y)(x² - xy + y² - xy)

= 3(x + y)(x - y)² > 0 (do x > y > 0)

=> b³ > a³

=> b > a (vì b;a > 0)

Answer 2