a: \(B=\frac{2\sqrt{x}+13}{\left(\sqrt{x}+2\right)\cdot\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}-2}{\sqrt{x}+2}\)
\(=\frac{2\sqrt{x}+13+\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{2\sqrt{x}+13+x+\sqrt{x}-6}{\left(\sqrt{x}+2\right)\cdot\left(\sqrt{x}+3\right)}=\frac{x+3\sqrt{x}+7}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
b: C=A-B
\(=\frac{2\sqrt{x}-1}{\sqrt{x}+3}-\frac{x+3\sqrt{x}+7}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+2\right)-\left(x+3\sqrt{x}+7\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+2\right)}=\frac{2x+3\sqrt{x}-2-x-3\sqrt{x}-7}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}-3}{\sqrt{x}+2}\)
C nguyên khi \(\sqrt{x}-3\) ⋮\(\sqrt{x}+2\)
=>\(\sqrt{x}+2-5\) ⋮\(\sqrt{x}+2\)
=>-5⋮\(\sqrt{x}+2\)
=>\(\sqrt{x}+2=5\) (Do \(\sqrt{x}+2\ge2\forall x\) thỏa mãn ĐKXĐ)
=>\(\sqrt{x}=3\)
=>x=9(nhận)
