Question

Cho x=\(\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\) voi a<0 ;b<0

a)CMR:\(x^2-4\ge0\)

b)Rut gon :\(\sqrt{x^2-4}\)

Answer 1

a) Xét \(x^2-4=\left(\sqrt{\frac{a}{b}}\right)^2+\left(\sqrt{\frac{b}{a}}\right)^2+2-4\)

\(=\left(\sqrt{\frac{a}{b}}\right)^2+\left(\sqrt{\frac{b}{a}}\right)^2-2=\left(\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\right)^2\ge0\)

b) \(\sqrt{x^2-4}=\sqrt{\left(\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\right)^2}=\left|\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\right|\)

Nếu a < b < 0 thì \(\sqrt{\frac{a}{b}}< \sqrt{\frac{b}{a}}\Rightarrow\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}< 0\Rightarrow\sqrt{x^2-4}=\sqrt{\frac{b}{a}}-\sqrt{\frac{a}{b}}\)Nếu b < a < 0 thì \(\sqrt{\frac{b}{a}}< \sqrt{\frac{a}{b}}\Rightarrow\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}>0\Rightarrow\sqrt{x^2-4}=\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\)

Answer 2

a) Vì a<0 , b<0 => \(\frac{a}{b}>0;\frac{b}{a}>0\Rightarrow\sqrt{\frac{a}{b}}>0;\sqrt{\frac{b}{a}}>0\)

Áp dụng bất đẳng thức cô si ta có:

 \(\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\ge2\sqrt{\sqrt{\frac{a}{b}}\cdot\sqrt{\frac{b}{a}}}=2\)

=> \(\left(\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\right)^2\ge4\)

Hay \(x^2\ge4\)