H24

B=\(\dfrac{\sqrt{x}+2}{\sqrt{x}-2}-\dfrac{3}{\sqrt{x}+2}-\dfrac{12}{x-4}\)

tìm x∈ Z để B>2

NL
14 tháng 9 2022 lúc 7:41

ĐKXĐ: \(x\ge0;x\ne4\)

\(B=\dfrac{\left(\sqrt{x}+2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{12}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{x+4\sqrt{x}+4-3\sqrt{x}+6-12}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{x+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

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

\(B>2\Rightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-2>0\Leftrightarrow\dfrac{3-\sqrt{x}}{\sqrt{x}-2}>0\)

\(\Rightarrow2< \sqrt{x}< 3\)

\(\Rightarrow4< x< 9\)

\(\Rightarrow x=\left\{5;6;7;8\right\}\)

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