Đáp án:
$\\$
`(3x × 1/5)^{200} + (2/5y + 4/7)^{1968} = 0`
Với mọi `x,y` có : \(\left\{ \begin{array}{l}(3x×\dfrac{1}{5})^{200} \geqslant 0\\(\dfrac{2}{5}y + \dfrac{4}{7})^{1968} \geqslant 0 \end{array} \right.\)
`↔ (3x×1/5)^{200} + (2/5y + 4/7)^{1968}` $\geqslant 0 ∀ x,y$
Dấu "`=`" xảy ra khi :
`↔` \(\left\{ \begin{array}{l}(3x×\dfrac{1}{5})^{200}=0\\(\dfrac{2}{5}y+\dfrac{4}{7})^{1968}=0\end{array} \right.\)
`↔` \(\left\{ \begin{array}{l}3x×\dfrac{1}{5}=0\\ \dfrac{2}{5}y+\dfrac{4}{7}=0\end{array} \right.\)
`↔` \(\left\{ \begin{array}{l}3x=0÷\dfrac{1}{5}\\ \dfrac{2}{5}y=0-\dfrac{4}{7}\end{array} \right.\)
`↔` \(\left\{ \begin{array}{l}3x=0\\ \dfrac{2}{5}y=\dfrac{-4}{7}\end{array} \right.\)
`↔` \(\left\{ \begin{array}{l}x=0÷3\\y=\dfrac{-4}{7}÷\dfrac{2}{5}\end{array} \right.\)
`↔` \(\left\{ \begin{array}{l}x=0\\y=\dfrac{-10}{7}\end{array} \right.\)
Vậy `(x;y) = (0;(-10)/7)`
