a) Thay x = 81 vào A ta có:
\(A=\dfrac{4\sqrt{81}}{\sqrt{81}-5}=\dfrac{4\cdot9}{9-5}=\dfrac{4\cdot9}{4}=9\)
b) \(B=\dfrac{\sqrt{x}-2}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+2}+\dfrac{5-2\sqrt{x}}{x+\sqrt{x}-2}\left(x\ne1;x\ge0\right)\)
\(B-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+2}+\dfrac{5-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(B=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\dfrac{5-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(B=\dfrac{x-4+\sqrt{x}-1+5-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(B=\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(B=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(B=\dfrac{\sqrt{x}}{\sqrt{x}+2}\)
c) \(\dfrac{A}{B}< 4\) khi
\(\dfrac{4\sqrt{x}}{\sqrt{x}-5}:\dfrac{\sqrt{x}}{\sqrt{x}+2}< 4\)
\(\Leftrightarrow\dfrac{4\left(\sqrt{x}+2\right)}{\sqrt{x}-5}< 4\)
\(\Leftrightarrow\dfrac{4\sqrt{x}+8-4\left(\sqrt{x}-4\right)}{\sqrt{x}-5}< 0\)
\(\Leftrightarrow\dfrac{24}{\sqrt{x}-5}< 0\)
\(\Leftrightarrow\sqrt{x}-5< 0\)
\(\Leftrightarrow x< 25\)
Kết hợp với đk:
\(0\le x< 5\)