Question

[Huỳnh Thị Thu Uyên]

Bài 2 : Tìm x,y biết rằng : ( 1/2x -5)²⁰ + ( y² -1/4)¹⁰ <  0

Answer 1

[Isolde Moria]

Ta có :

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

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

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

(+) \(\frac{1}{2x}-5=0\)

\(\Rightarrow x=\frac{1}{10}\)

(+) \(y^2-\frac{1}{4}=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}y=\frac{1}{2}\\y=-\frac{1}{2}\end{array}\right.\)

Vậy \(\left(x;y\right)\in\left\{\left(\frac{1}{10};\frac{1}{2}\right);\left(\frac{1}{10};-\frac{1}{2}\right)\right\}\)

Answer 2

[soyeon_Tiểubàng giải]

Do \(\left(\frac{1}{2}x-5\right)^{20}\ge0;\left(y^2-\frac{1}{4}\right)^{10}\ge0\)

=> \(\left(\frac{1}{2}x-5\right)^{20}+\left(y^2-\frac{1}{4}\right)^{10}\ge0\)

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

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

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