\(\sqrt{29-x}+\sqrt{x+3}=x^2-26x+177\left(1\right)\)
ĐK -3 =<x =<29
Với mọi a,b >=0 ta có:
\(\left(a-b\right)^2\ge0\)
\(\Leftrightarrow2\left(a^2+b^2\right)\ge2ab\)
\(\Leftrightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)
\(\Leftrightarrow a+b\le\sqrt{2\left(a^2+b^2\right)}\)
Thay \(a=\sqrt{29-x};b=\sqrt{x+3}\)ta có:
\(\sqrt{29-x}+\sqrt{x+3}\le\sqrt{2\left(29-x+x+3\right)}=8\)
\(x^2-26x+177=\left(x-13\right)^2+8\ge8\)
\(\Rightarrow\sqrt{29-x}+\sqrt{x+3}\le x^2-26x+177\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}\sqrt{29-x}=\sqrt{x+3}\\x-13=0\end{cases}\Leftrightarrow x=13}\)
Do đó (1) <=> x=13 (tm)
