Nhận xet: y bằng tổng 2 ân => y ≥ 0
\(y^2=x-1+4-x+2\sqrt{\left(x-1\right)\left(4-x\right)}=3+2\sqrt{\left(x-1\right)\left(4-x\right)}\)
Vì \(2\sqrt{\left(x-1\right)\left(4-x\right)}\ge0\)
=> \(y^2\ge3\) mà y ≥ 0
=> y ≥ \(\sqrt{3}\). Dấu "=" xảy ra <=> x = 1 hoặc x = 4
Lại có: \(2\sqrt{\left(x-1\right)\left(4-x\right)}\le2.\dfrac{x-1+4-x}{2}=3\)
=> \(y^2\le6\)
Mà y ≥ 0
=> y ≤ \(\sqrt{6}\)
Dấu "=" xảy ra <=> x = \(\dfrac{5}{2}\)
ĐKXĐ: \(1\le x\le4\)
-Min:
Với x > 0, Áp dụng BĐT :\(\sqrt{a}+\sqrt{b}\ge\sqrt{a+b}\)
\(\Rightarrow y=\sqrt{x-1}+\sqrt{4-x}\ge\sqrt{3}\)
\(''=''\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)
-Max:
\(y^2=\left(\sqrt{x-1}+\sqrt{4-x}\right)^2\)\(=3+2\sqrt{\left(x-1\right)\left(4-x\right)}\)
\(y^2\le3+2.\dfrac{x-1+4-x}{2}=6\)
\(y\le\sqrt{6}\)
\(''=''\Leftrightarrow x=\dfrac{5}{2}\)
