Question

Answer 1

Bài 3:

a: \(B=\dfrac{\sqrt{a}}{\sqrt{a}-3}-\dfrac{3}{\sqrt{a}+3}-\dfrac{a-2}{a-9}\)

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

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

\(=\dfrac{a+3\sqrt{a}-3\sqrt{a}+9-a+2}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}=\dfrac{11}{a-9}\)

b: Để B là số nguyên thì \(11⋮a-9\)

=>\(a-9\in\left\{1;-1;11;-11\right\}\)

=>\(a\in\left\{10;8;20;-2\right\}\)

kết hợp ĐKXĐ, ta được: \(a\in\left\{10;8;20\right\}\)