Question

Rút gọn gọn

Tìm a để B<1

\(B=\dfrac{\sqrt a+3}{2\sqrt a-6}-\dfrac{3-\sqrt a}{2\sqrt a+6}\)

Answer 1

ĐK:\(a\ge0;a\ne9\)

\(B=\frac{\sqrt{a}+3}{2\left(\sqrt{a}-3\right)}+\frac{\sqrt{a}-3}{2\left(\sqrt{a}+3\right)}\)

\(=\frac{2\left(\sqrt{a}+3\right)^2+2\left(\sqrt{a}-3\right)^2}{4\left(a-9\right)}\)\(=\frac{a+9}{a-9}\)

Answer 2

\(B=\dfrac{a+6\sqrt{a}+9+a-6\sqrt{a}+9}{2\left(a-9\right)}=\dfrac{2a+18}{2a-18}\)

Để B<1 thì B-1<0

=>(2a+18-2a+18)/(2a-18)<0

=>2a-18<0

=>0<=a<9