Question

Tìm x,y thoả mãn :\

( 3x-5/9)^2018 + ( 3y+0,4/3)^2020 = 0

Answer 1

\(\left(\dfrac{3x-5}{9}\right)^{2018}>=0\forall x\)

\(\left(\dfrac{3y+0,4}{3}\right)^{2020}>=0\forall y\)

Do đó: \(\left(\dfrac{3x-5}{9}\right)^{2018}+\left(\dfrac{3y+0,4}{3}\right)^{2020}>=0\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}\dfrac{3x-5}{9}=0\\\dfrac{3y+0,4}{3}=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x-5=0\\3y+0,4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{3}\\y=-\dfrac{0.4}{3}=-\dfrac{2}{15}\end{matrix}\right.\)