Question

\(\left(\frac{1}{2}x-5\right)^{20}+\left(y^2-\frac{1}{4}\right)^{10}\le0\)

Tìm x và y 

Answer 1

Vì \(\hept{\begin{cases}\left(\frac{1}{2}x-5\right)^{20}\ge0\\\left(y^2-\frac{1}{4}\right)^{10}\ge0\end{cases}\Rightarrow\left(\frac{1}{2}x-5\right)^{20}+\left(y^2-\frac{1}{4}\right)^{10}}\ge0\)

Theo đề bài:

 \(\left(\frac{1}{2}x-5\right)^{20}+\left(y^2-\frac{1}{4}\right)^{10}\le0\)

=> \(\left(\frac{1}{2}x-5\right)^{20}+\left(y^2-\frac{1}{4}\right)^{10}=0\)

<=>\(\hept{\begin{cases}\left(\frac{1}{2}x-5\right)^{20}=0\\\left(y^2-\frac{1}{4}\right)^{10}=0\end{cases}}\)

<=>\(\hept{\begin{cases}\frac{1}{2}x-5=0\\y^2-\frac{1}{4}=0\end{cases}}\)

<=>\(\hept{\begin{cases}\frac{1}{2}x=5\\y^2=\frac{1}{4}\end{cases}}\)

<=>\(x=10\) và \(y=-\frac{1}{4}\) hoặc \(y=\frac{1}{4}\)

Vậy ...

Answer 2

thanks