Question

Tìm x , y biết

a) x² + = 0

b) (\(\frac{1}{2}\)x - 5)²⁰ + (y² - \(\frac{1}{4}\))¹⁰ nhỏ hơn hoặc = 0

Answer 1

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

Ta thấy: \(\left\{\begin{matrix}x^2\ge0\\\left(y-\frac{1}{10}\right)^4\ge0\end{matrix}\right.\)

\(\Rightarrow x^2+\left(y-\frac{1}{10}\right)^4\ge0\)

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

Xảy ra khi \(\left\{\begin{matrix}x^2=0\\\left(y-\frac{1}{10}\right)^4=0\end{matrix}\right.\)\(\Leftrightarrow\left\{\begin{matrix}x=0\\y=\frac{1}{10}\end{matrix}\right.\)

Answer 2

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

Ta thấy: \(\left\{\begin{matrix}\left(x-5\right)^{20}\ge0\\\left(y^2-\frac{1}{4}\right)^{10}\ge0\end{matrix}\right.\)

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

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

Suy ra \(\left\{\begin{matrix}\left(x-5\right)^{20}=0\\\left(y^2-\frac{1}{4}\right)^{10}=0\end{matrix}\right.\)\(\Rightarrow\left\{\begin{matrix}x-5=0\\y^2-\frac{1}{4}=0\end{matrix}\right.\)\(\Rightarrow\left\{\begin{matrix}x=5\\y=\pm\frac{1}{2}\end{matrix}\right.\)

Answer 3

Nguyễn Huy Tú

Answer 4

đề câu a) thiếu