lấy tạm 1 ví dụ thôi nhé!
Giải phương trình \(\sqrt{x-1}+\sqrt{3-x}+4x\sqrt{2x}=x^3+10\)
ĐK: \(1\le x\le3\)
\(\sqrt{x-1}=\frac{1}{2}.2.1.\sqrt{x-1}\le\frac{1}{2}\left(1+x-1\right)=\frac{1}{2}.x\) ( Cô - si )
\(\sqrt{3-x}=\frac{1}{2}.2.1.\sqrt{3-x}\le\frac{1}{2}\left(1+3-x\right)=2-\frac{1}{2}x\)
\(4x\sqrt{2x}=2.2.\sqrt{2}.\sqrt{x^3}\le8+x^3\)
\(\Rightarrow\sqrt{x-1}+\sqrt{3-x}+4x\sqrt{2x}\le x^3+10\)
Dấu "=" khi \(\hept{\begin{cases}1=\sqrt{x-1}\\1=\sqrt{3-x}\\2\sqrt{2}=\sqrt{x^3}\end{cases}\Leftrightarrow x=2}\left(tmđk\right)\)
