Question

cho x,y thỏa mãn \(\left(\dfrac{1}{3}-2x\right)^{2018}+\left(3y-x\right)^{2020}\le0\)

Chứng tỏ:\(\dfrac{1}{x}+\dfrac{1}{y}=24\)

Help Mn ơi lm giúp mk vs chiều nộp rồi

Answer 1

Ta có: \(\left(\dfrac{1}{3}-2x\right)^{2018}\ge0\forall x\);

\(\left(3y-x\right)^{2020}\ge0\forall x;y\)

=> \(\left(\dfrac{1}{3}-2x\right)^{2018}+\left(3y-x\right)^{2020}\ge0\)

mà theo đề thì:\(\left(\dfrac{1}{3}-2x\right)^{2018}+\left(3y-x\right)^{2020}\le0\)

=> Dấu ''='' xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}\dfrac{1}{3}-2x=0\\3y-x=0\end{matrix}\right.\)

Ta có: \(\dfrac{1}{3}-2x=0\Rightarrow x=\dfrac{1}{6}\);

\(3y-x=0\Leftrightarrow3y-\dfrac{1}{6}=0\Leftrightarrow3y=\dfrac{1}{6}\Leftrightarrow y=\dfrac{1}{18}\)

=> \(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{\dfrac{1}{6}}+\dfrac{1}{\dfrac{1}{18}}=6+18=24\left(đpcm\right)\)