Question

Answer 1

a) Ta có: \(S=\left(\dfrac{\sqrt{14}-\sqrt{7}}{\sqrt{2}-1}+\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)

\(=\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)

=7-5=2

b) Ta có: \(T=\sqrt{\left(a-3\right)^2}+\left(a-9\right)\)

\(=3-a+a-9\)

=-6

Answer 2

Lời giải:
\(S=\left[\frac{\sqrt{7}(\sqrt{2}-1)}{\sqrt{2}-1}+\frac{\sqrt{5}(\sqrt{3}-1)}{\sqrt{3}-1}\right].(\sqrt{7}-\sqrt{5})\)

\(=(\sqrt{7}+\sqrt{5})(\sqrt{7}-\sqrt{5})=7-5=2\)

\(T=\sqrt{(a-3)^2}+(a-9)=|a-3|+(a-9)\)

\(=3-a+a-9=-6\) (do $a< 3$ thì $|a-3|=3-a$)