a, Đk: \(2\le x\le4\)
Áp dụng Caushy:
\(A^2=\left(\sqrt{x-2}+\sqrt{4-x}\right)^2\le\left[\left(\sqrt{x-2}\right)^2+\left(\sqrt{4-x}\right)^2\right]^2=\left(x-2+4-x\right)^2=4\)
\(\Rightarrow0< A\le2\)
Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x-2}=\sqrt{4-x}\Leftrightarrow x=3\)
- Ta có bđt: \(\sqrt{x}+\sqrt{y}\ge\sqrt{x+y}\left(x,y\ge0\right)\)
Dấu "=" xảy ra \(\Leftrightarrow\left[{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
Áp dụng: \(A=\sqrt{x-2}+\sqrt{4-x}\ge\sqrt{x-2+4-x}=\sqrt{2}\)
Dấu "=" xảy ra \(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\4-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)
Vậy \(MaxA=2;MinA=\sqrt{2}\)
b, Tương tự.
Bài 2:
- Bđt Bunhiacopxki cho hai số:
\(\left(ax+by\right)^2\le\left(a^2+b^2\right)\left(x^2+y^2\right)\)
Dấu "=" xảy ra \(\Leftrightarrow\dfrac{a}{x}=\dfrac{b}{y}\) (Quy ước: nếu \(x=0,y=0\) thì \(a=b=0\) và ngược lại).
b, Áp dụng Bunhiacopxki:
\(\left(x^2+y^2\right)\left(1^2+2^2\right)\ge\left(x.1+y.2\right)^2\)
\(\Leftrightarrow\left(x^2+y^2\right).5\ge\left(x+2y\right)^2\)
\(\Leftrightarrow\left(x+2y\right)^2\le25\)
\(\Leftrightarrow0< x+2y\le5\)
\(MinB=5\Leftrightarrow\dfrac{x}{1}=\dfrac{y}{2};x^2+y^2=5\Leftrightarrow x=1;y=2\)
a,c: Tương tự.
