Question

1. So sánh: \(\sqrt{3}\) và \(5-\sqrt{8}\)

2. Tìm Min, Max

a/ x = \(\sqrt{x^2-2x+5}\)

b/ y = \(\sqrt{\dfrac{x^2}{4}-\dfrac{x}{6}+1}\)

Answer 1

\(x=\sqrt{x^2-2x+5}=\sqrt{x^2-2x+1+4}\\ =\sqrt{\left(x-1\right)^2+4}\ge\sqrt{4}=2\)

dấu "=" xảy ra khi x=1

vậy min x=2 khi x=1

Answer 2

\(y=\sqrt{\dfrac{x^2}{4}-\dfrac{x}{6}+1}=\sqrt{\left(\dfrac{x}{2}\right)^2-2.\dfrac{x}{2}.\dfrac{1}{6}+\dfrac{1}{36}+\dfrac{35}{36}}\\ =\sqrt{\left(\dfrac{x}{2}-\dfrac{1}{6}\right)^2+\dfrac{35}{36}}\ge\sqrt{\dfrac{35}{36}}\)

dấu "=" xảy ra khi \(\dfrac{x}{2}-\dfrac{1}{6}=0\Rightarrow x=\dfrac{1}{3}\)

vậy min y =\(\sqrt{\dfrac{35}{36}}\) tại \(x=\dfrac{1}{3}\)

Answer 3

1

\(\left(5-\sqrt{8}\right)^2=33-20\sqrt{2}>3\\ \Rightarrow5-\sqrt{8}>\sqrt{3}\)