Ta có:/x-3/^2014>=0;/6+2y/^2015>=0

=>/x-3/^2014+/6+2y/^2015>=0

mà theo đề bài, /x-3/^2014+/6+2y/^2015<=0

=>/x-3/^2014=/6+2y/^2015=0

=>/x-3/=0;       /6+2y/=0

=>x-3=0           =>6+2y=0

=>x=3              =>2y=-6=>y=-3

vậy x=3; y=-3

3 hoặc -3 đúng zồi đó

3;-3 đúng 1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000%

x=3

y=-3

T..i..c..k mk nha

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

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

Mà đề lại cho : \(\left|x-3\right|^{2014}+\left|6+2y\right|^{2015}\le0\Rightarrow\left|x-3\right|^{2014}=0;\left|6+2y\right|^{2015}=0\)

\(\Rightarrow x-3=0;6+2y=0\Rightarrow x=3;y=-3\)

x=3 ; y=3

x=y=3

Ta có:\(\hept{\begin{cases}\left|x-3\right|\ge0\\\left|6+2y\right|\ge0\end{cases}\Rightarrow\hept{\begin{cases}\left|x-3\right|^{2014}\ge0\\\left|6+2y\right|^{2015}\ge0\end{cases}\Rightarrow}\left|x-3\right|^{2014}+\left|6+2y\right|^{2015}\ge0}\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}\left|x-3\right|^{2014}=0\\\left|6-2y\right|^{2015}=0\end{cases}\Rightarrow\hept{\begin{cases}x=3\\y=3\end{cases}}}\)

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

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

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

=> \(x-3=6+2y=0\)

=> \(x=3;y=-3\).

Ta có:\(y=\frac{1}{x^2+\sqrt{x}}\)

ket qua x=3 ; y=-3

Vì |x - 3|²⁰¹⁴ ≥ 0 ; |6 + 2y|²⁰¹⁵ ≥ 0

=> |x - 3|²⁰¹⁴ + |6 + 2y|²⁰¹⁵ ≥ 0

Mà để |x - 3|²⁰¹⁴ + |6 + 2y|²⁰¹⁵ ≤ 0 <=> |x - 3|²⁰¹⁴ = 0 ; |6 + 2y|²⁰¹⁵ = 0

=> x = 3 và y = - 3

Vậy  x = 3 và y = - 3