Ta có \(\left(\sqrt{101}-\sqrt{99}\right)^2=101+99-\sqrt{39996}=200-\sqrt{39996}=0,01+199,99-\sqrt{39996}\)\(\left(0,1\right)^2=0,01\)
Ta có \(39996,0001>39996\Rightarrow\sqrt{39996,0001}>\sqrt{39996}\Rightarrow199,99>\sqrt{39996}\Rightarrow199,99-\sqrt{39996}>0\Rightarrow0,01+199,99-\sqrt{39996}>0,01\Rightarrow\sqrt{0,01+199,99-\sqrt{39996}}>\sqrt{0,01}\Rightarrow\sqrt{101}-\sqrt{99}>0,1\)
Ta có:
\(\sqrt{100}+\sqrt{99}\\ =\sqrt{100+1}+\sqrt{100-1}< 2\sqrt{100}=20\\ \Rightarrow\sqrt{101}+\sqrt{99}< 20\\ \Leftrightarrow\dfrac{\sqrt{101}+\sqrt{99}}{2}< 10\\ \Rightarrow\dfrac{\sqrt{101}+99}{\left(\sqrt{101}+\sqrt{99}\right)\left(\sqrt{101}-\sqrt{99}\right)}< 0\\ \Leftrightarrow\dfrac{1}{\sqrt{101}-\sqrt{99}}< 10\\ \Leftrightarrow\sqrt{101}-\sqrt{99}>\dfrac{1}{10}=0,1\)
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