a: Thay x=9 vào A, ta được:
\(A=\frac{\sqrt9+1}{\sqrt9+2}=\frac{3+1}{3+2}=\frac45\)
b: \(B=\frac{\sqrt{x}-11}{x-\sqrt{x}-2}-\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{2\sqrt{x}-1}{\sqrt{x}-2}\)
\(=\frac{\sqrt{x}-11}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{2\sqrt{x}-1}{\sqrt{x}-2}\)
\(=\frac{\sqrt{x}-11-\sqrt{x}\left(\sqrt{x}-2\right)+\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}-11-x+2\sqrt{x}+2x+2\sqrt{x}-\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}=\frac{x+4\sqrt{x}-12}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(\sqrt{x}+6\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}+6}{\sqrt{x}+1}\)
c: Đặt P=A* B
\(=\frac{\sqrt{x}+6}{\sqrt{x}+1}\cdot\frac{\sqrt{x}+1}{\sqrt{x}+2}=\frac{\sqrt{x}+6}{\sqrt{x}+2}\)
đê P là số nguyên thì \(\sqrt{x}+6\) ⋮\(\sqrt{x}+2\)
=>\(\sqrt{x}+2+4\) ⋮\(\sqrt{x}+2\)
=>4⋮\(\sqrt{x}+2\)
=>\(\sqrt{x}+2\in\left\lbrace2;4\right\rbrace\)
=>\(\sqrt{x}\in\left\lbrace0;2\right\rbrace\)
=>x∈{0;4}
Kết hợp ĐKXĐ, ta được: x=0
