\(1,\) Khi \(x=9\) thì \(A=\dfrac{x+\sqrt{x}+4}{\sqrt{x}-2}=\dfrac{9+\sqrt{9}+4}{\sqrt{9}-2}=16\)
\(2,\) Ta có \(B=\dfrac{3x-4}{x-2\sqrt{x}}-\dfrac{\sqrt{x}+2}{\sqrt{x}}+\dfrac{\sqrt{x}-1}{2-\sqrt{x}}\left(x>0;x\ne4\right)\)
\(B=\dfrac{3x-4-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\\ B=\dfrac{3x-4-x+4-x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\\ B=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\)
1: Thay x=9 vào A,ta được:
\(A=\dfrac{9+3+4}{3-2}=16\)
2: Ta có: \(B=\dfrac{3x-4}{x-2\sqrt{x}}-\dfrac{\sqrt{x}+2}{\sqrt{x}}+\dfrac{\sqrt{x}-1}{2-\sqrt{x}}\)
\(=\dfrac{3x-4-\left(x-4\right)-\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3x-4-x+4-x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\)
