Question

[Unhappy]

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Answer 1

[Đoàn Đức Hà]

Bài I: 

a) Khi \(x=9\): 

\(A=\frac{\sqrt{9}+4}{\sqrt{9}-1}=\frac{3+4}{3-1}=\frac{7}{2}\)

b) \(B=\frac{3\sqrt{x}+1}{x+2\sqrt{x}-3}-\frac{2}{\sqrt{x}+3}=\frac{3\sqrt{x}+1}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\frac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{3\sqrt{x}+1-2\sqrt{x}+2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\frac{1}{\sqrt{x}-1}\)

c) \(\frac{A}{B}=\frac{\sqrt{x}+4}{\sqrt{x}-1}\div\frac{1}{\sqrt{x}-1}=\sqrt{x}+4\ge\frac{x}{4}+5\)

\(\Leftrightarrow x-4\sqrt{x}+4\le0\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)^2\le0\)

\(\Leftrightarrow\sqrt{x}-2=0\)

\(\Leftrightarrow x=4\).