Tìm Min : \(A=\sqrt{21+4a-a^2}+\sqrt{10+3a-a^2}\)
@Nguyễn Việt Lâm, @Akai Haruma
giúp em với ạ! Em cảm ơn nhiều!
ĐKXĐ: \(-2\le a\le5\)
\(\Rightarrow\left\{{}\begin{matrix}\left(a+2\right)\left(5-a\right)\ge0\\\left(a+2\right)\left(6-a\right)\ge0\end{matrix}\right.\)
\(A=\sqrt{9+\left(a+2\right)\left(6-a\right)}+\sqrt{\left(a+2\right)\left(5-a\right)}\ge\sqrt{9+\left(a+2\right)\left(6-a\right)}\ge\sqrt{9}=3\)
Dấu "=" xảy ra khi \(a=-2\)
ĐKXĐ: \(-2\le a\le5\)
\(A=\frac{11+a}{\sqrt{21+4a-a^2}+\sqrt{10+3a-a^2}}>0\)
\(A^2=-2a^2+7a+31-2\sqrt{\left(a+3\right)\left(7-a\right)\left(a+2\right)\left(5-a\right)}\)
\(A^2=-2a^2+7a+31-2\sqrt{\left(-a^2+2a+15\right)\left(-a^2+5a+14\right)}\)
\(A^2\ge-2a^2+7a+31-\left(-a^2+2a+15-a^2+5a+14\right)\)
\(A^2\ge-2a^2+7a+31+2a^2-7a-29\)
\(A^2\ge2\Rightarrow A\ge\sqrt{2}\)
\(A_{min}=\sqrt{2}\) khi \(-a^2+2a+15=-a^2+5a+14\Leftrightarrow a=\frac{1}{3}\)
