a) Ta có:

\(6x^2+5y^2=74\)

\(\Rightarrow6\left(x^2-4\right)=5\left(10-y^2\right)\) (1)

Từ (1) \(\Rightarrow6\left(x^2-4\right)⋮5\) và (5,6)=1

\(\Rightarrow x^2-4⋮5\Rightarrow x^2=5k+4\left(k\in N\right)\)

Thay \(x^2-4=5k\) vào (1) ta có:

\(\Rightarrow y^2=10-6k\)

Vì\(\left\{{}\begin{matrix}x^2>0\\y^2>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}5k+4>0\\10k-4>0\end{matrix}\right.\)

\(\Rightarrow-\dfrac{4}{5}< k< \dfrac{5}{3}\Rightarrow\left[{}\begin{matrix}k=0\\k=1\end{matrix}\right.\)

(+) Nếu k = 0 \(\Rightarrow y^2=10\) (loại)

(+) Nếu k = 1 \(\Rightarrow\left\{{}\begin{matrix}x^2=9\\y^2=4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\pm3\\y=\pm4\end{matrix}\right.\)

Vậy (x,y) \(\in\left\{\left(3,2\right);\left(-3,-2\right)\right\}\)

Vì \(\hept{\begin{cases}\left(2x-1\right)^{2010}\ge0\\\left(y-\frac{2}{5}\right)^{2010}\ge0\\\left|x+y-z\right|\ge0\end{cases}\forall x,y,z}\)

\(\Rightarrow\left(2x-1\right)^{2010}+\left(y-\frac{2}{5}\right)^{2010}+\left|x+y-z\right|\ge0\)

Mà \(\left(2x-1\right)^{2010}+\left(y-\frac{2}{5}\right)^{2010}+\left|x+y-z\right|=0\)

\(\Leftrightarrow\hept{\begin{cases}\left(2x-1\right)^{2010}=0\\\left(y-\frac{2}{5}\right)^{2010}=0\\\left|x+y-z\right|=0\end{cases}\Leftrightarrow\hept{\begin{cases}2x-1=0\\y-\frac{2}{5}=0\\x+y-z=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{2}{5}\\z=\frac{9}{10}\end{cases}}}\)

Vậy...