Question

[Phương Thảo Lâm]

Tìm x;y biết \(\text{|}x-3\text{|}^{2014}+\text{|}6+2y\text{|}^{2015}\le0\)

Answer 1

[Minh Hiền]

Ta có: \(\left|x-3\right|^{2014}\ge0;\left|6+2y\right|^{2015}\ge0\)

\(\Rightarrow\left|x-3\right|^{2014}+\left|6+2y\right|^{2015}\ge0\)

Mà theo đề: \(\left|x-3\right|^{2014}+\left|6+2y\right|^{2015}\le0\)

=> \(\left|x-3\right|^{2014}+\left|6+2y\right|^{2015}=0\)

=> \(\left|x-3\right|=\left|6+2y\right|=0\)

=> \(x-3=6+2y=0\)

=> \(x=3;y=-3\).

Answer 2

[Nguyễn Mạnh Trung]

Ta có:\(y=\frac{1}{x^2+\sqrt{x}}\)