\(\sqrt{a}+\sqrt{b}\le\sqrt{2\left(a+b\right)}\Leftrightarrow\left(\sqrt{a}+\sqrt{b}\right)^2\le\sqrt{2\left(a+b\right)}\left(vì:\sqrt{a}+\sqrt{b},\sqrt{2\left(a+b\right)}\ge0\right)\Leftrightarrow a+2\sqrt{ab}+b\le2a+2b\Leftrightarrow2\sqrt{ab}\le a+b\Leftrightarrow a-2\sqrt{ab}+b\ge0\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\left(luondung\right)\Rightarrow\sqrt{a}+\sqrt{b}\le\sqrt{2\left(a+b\right)}\) áp dungj ta đc: \(M=\sqrt{x-5}+\sqrt{7-x}\le\sqrt{2\left(x-5+7-x\right)}=\sqrt{4}=2\)
Nên GTLN cua M là 2 dâus "=" xayr ra khi:x=6
\(M^2=x-5+7-x+2\sqrt{\left(x-5\right)\left(7-x\right)}=2+2\sqrt{\left(x-5\right)\left(7-x\right)}\) \(Mà:\sqrt{\left(x-5\right)\left(7-x\right)}\ge0\Rightarrow2\sqrt{\left(x-5\right)\left(7-x\right)}\ge0\Rightarrow2+2\sqrt{\left(x-5\right)\left(7-x\right)}\ge2\Rightarrow M^2\ge2\) \(Mà:\left\{{}\begin{matrix}\sqrt{x-5}\ge0\\\sqrt{7-x}\ge0\end{matrix}\right.\Rightarrow\sqrt{x-5}+\sqrt{7-x}\ge0\Rightarrow M\ge0\)
\(M^2\ge2\Rightarrow M\ge\sqrt{2}\left(do:M\ge0\right)\)
Nên GTNN cua M là: \(\sqrt{2}\)
Dâus "=" xayr ra \(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=7\end{matrix}\right.\)
Đính chính lại cho mk dongf 1 là:
\(\sqrt{a}+\sqrt{b}\le\sqrt{2\left(a+b\right)}\Leftrightarrow\left(\sqrt{a}+\sqrt{b}\right)^2\le2a+2b\left(vì:\sqrt{a}+\sqrt{b},\sqrt{2\left(a+b\right)}\ge0\right)\)
nha
