Question

[Nguyễn Thị Thanh Trang]

Cho \(2018x^3=2019y^3=2020z^3\) và \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=8\)

Tính giá trị biểu thức: \(B=\frac{\sqrt[3]{2018x^2+2019y^2+2020z^2}}{\sqrt[3]{2018}+\sqrt[3]{2019}+\sqrt[3]{2020}}\)

Answer 1

[Akai Haruma]

Lời giải:

Đặt mẫu số của $B$ là $M$.

Từ \(2018x^3=2019y^3=2020z^3\)

\(\Rightarrow \sqrt[3]{2018}x=\sqrt[3]{2019}y=\sqrt[3]{2020}z=\frac{\sqrt[3]{2018}}{\frac{1}{x}}=\frac{\sqrt[3]{2019}}{\frac{1}{y}}=\frac{\sqrt[3]{2020}}{\frac{1}{z}}=\frac{\sqrt[3]{2018}+\sqrt[3]{2019}+\sqrt[3]{2020}}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}\)

\(=\frac{\sqrt[3]{2018}+\sqrt[3]{2019}+\sqrt[3]{2020}}{8}=\frac{M}{8}\)

\(\Rightarrow \left\{\begin{matrix} x=\frac{M}{8\sqrt[3]{2018}}\\ y=\frac{M}{8\sqrt[3]{2019}}\\ z=\frac{M}{8\sqrt[3]{2020}}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} 2018x^2=\frac{\sqrt[3]{2018}M^2}{64}\\ 2019y^2=\frac{\sqrt[3]{2019}M^2}{64}\\ 2020z^2=\frac{\sqrt[3]{2020}M^2}{64}\end{matrix}\right.\)

\(\Rightarrow 2018x^2+2019y^2+2020z^2=\frac{M^2(\sqrt[3]{2018}+\sqrt[3]{2019}+\sqrt[3]{2020})}{64}=\frac{M^3}{64}\)

\(\Rightarrow B=\frac{\sqrt[3]{\frac{M^3}{64}}}{M}=\frac{M}{4M}=\frac{1}{4}\)