Question

[Đăng Khoa Nguyễn]

A=x^2+x+y^2-10y+30

ai trả lời nhanh mình tick nhé

Answer 1

[Đinh quang hiệp]

\(A=\left(x^2+x+\frac{1}{4}\right)+\left(y^2-10y+25\right)+\frac{19}{4}\)

\(=\left(x^2+2\cdot\frac{1}{2}x+\left(\frac{1}{2}\right)^2\right)+\left(y^2-2\cdot5y+5^2\right)+\frac{19}{4}=\left(x+\frac{1}{2}\right)^2+\left(y-5\right)^2+\frac{19}{4}>=\frac{19}{4}\)

dấu = xảy ra khi \(\left(x+\frac{1}{2}\right)^2=0\Rightarrow x+\frac{1}{2}=0\Rightarrow x=-\frac{1}{2}\)

                        \(\left(y-5\right)^2=0\Rightarrow y-5=0\Rightarrow y=5\)

vậy min A là \(\frac{19}{4}\)khi \(x=-\frac{1}{2};y=5\)

Answer 2

[_Guiltykamikk_]

( đề là tìm gtnn à ??? )

\(A=x^2+x+y^2-10y+30\)

\(A=\left(x^2+x+\frac{1}{4}\right)+\left(y^2-10y+25\right)+\frac{19}{4}\)

\(A=\left(x+\frac{1}{2}\right)^2+\left(y-5\right)^2+\frac{19}{4}\)

Mà  \(\left(x+\frac{1}{2}\right)^2\ge0\)

       \(\left(y-5\right)^2\ge0\)

\(\Rightarrow A\ge\frac{19}{4}\)

Dấu " = " xảy ra khi :

\(\hept{\begin{cases}x+\frac{1}{2}=0\\y-5=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{1}{2}\\y=5\end{cases}}\)

Vậy  \(A_{Min}=\frac{19}{4}\Leftrightarrow\left(x;y\right)=\left(-\frac{1}{2};5\right)\)