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Lời giải:
Ta thấy:
\(\frac{1}{2}\text{VP}=\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{3}}+...+\frac{1}{2\sqrt{100}}\)
\(> \frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{100}+\sqrt{101}}\)
Mà:
\(\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{100}+\sqrt{101}}=\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{2}+\sqrt{3}(\sqrt{3}-\sqrt{2})}+\frac{\sqrt{4}-\sqrt{3}}{(\sqrt{3}+\sqrt{4})(\sqrt{4}-\sqrt{3)}}+...+\frac{\sqrt{101}-\sqrt{100}}{(\sqrt{100}+\sqrt{101})(\sqrt{101}-\sqrt{100})}\)
\(=\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{3}}{4-3}+...+\frac{\sqrt{101}-\sqrt{100}}{101-100}\)
\(=\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{101}-\sqrt{100}\)
\(=\sqrt{101}-\sqrt{2}\)
Do đó: \(\frac{1}{2}\text{VP}> \sqrt{101}-\sqrt{2}\Rightarrow \text{VP}>2(\sqrt{101}-\sqrt{2})> 17\) (đpcm)
Ta có: \(\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}=2.\left(\dfrac{1}{\sqrt{2}+\sqrt{2}}+\dfrac{1}{\sqrt{3}+\sqrt{3}}+...+\dfrac{1}{\sqrt{100}+\sqrt{100}}\right)\) (1)
\(\left(1\right)< 2.\left(\dfrac{1}{\sqrt{2}+\sqrt{1}}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+...+\dfrac{1}{\sqrt{100}+\sqrt{99}}\right)\)\(=2\left(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}\right)\)\(=2\left(-\sqrt{1}+\sqrt{100}\right)=2\left(-1+10\right)=18\)
Vậy:...
Đặt A=\(\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{4}}+...+\dfrac{1}{\sqrt{100}}\)
\(\Leftrightarrow A=\dfrac{2}{2\sqrt{2}}+\dfrac{2}{2\sqrt{3}}+....+\dfrac{2}{2\sqrt{100}}\)
\(\Leftrightarrow A=\dfrac{2}{\sqrt{2}+\sqrt{2}}+\dfrac{2}{\sqrt{3}+\sqrt{3}}+....+\dfrac{2}{\sqrt{99}+\sqrt{99}}+\dfrac{2}{\sqrt{100}+\sqrt{100}}\)
\(\Leftrightarrow A=2\left(\dfrac{1}{\sqrt{2}+\sqrt{2}}+\dfrac{1}{\sqrt{3}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{99}}+\dfrac{1}{\sqrt{100}+\sqrt{100}}\right)\)
Ta có:
\(\dfrac{1}{\sqrt{2}+\sqrt{2}}< \dfrac{1}{1+\sqrt{2}};\dfrac{1}{\sqrt{3}+\sqrt{3}}< \dfrac{1}{\sqrt{2}+\sqrt{3}}\)
Tường tự, ta có:
\(\dfrac{A}{2}< \dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)
\(A< 2\left(\dfrac{1-\sqrt{2}}{-1}+\dfrac{\sqrt{2}-\sqrt{3}}{-1}+\dfrac{\sqrt{99}-\sqrt{100}}{-1}\right)\)
\(A< -2\left(1-\sqrt{2}+\sqrt{2}-\sqrt{3}+...-\sqrt{99}+\sqrt{99}-\sqrt{100}\right)\)
\(A< -2\left(1-\sqrt{100}\right)\)
\(A< 18\)
Vậy\(\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{4}}+...+\dfrac{1}{\sqrt{100}}< 18\)
VT tương đương với \(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)
\(=\dfrac{\sqrt{1}-\sqrt{2}}{1-2}+\dfrac{\sqrt{2}-\sqrt{3}}{2-3}+...+\dfrac{\sqrt{99}-\sqrt{100}}{99-100}\)
\(=\sqrt{100}-\sqrt{99}+\sqrt{99}-....-\sqrt{3}+\sqrt{3}-\sqrt{2}+\sqrt{2}-\sqrt{1}\) (kiểu do mẫu số nó có kết quả âm nên đảo lại phép)
\(=10-1=9=VP\)
a)=\(\dfrac{3\sqrt{6}}{2}+\dfrac{2\sqrt{6}}{3}-\dfrac{4\sqrt{6}}{2}\)
\(=\dfrac{2\sqrt{6}}{3}-\dfrac{\sqrt{6}}{2} \)
=\(\dfrac{4\sqrt{6}}{6}-\dfrac{3\sqrt{6}}{6}=\dfrac{\sqrt[]{6}}{6}\)
b)\(\dfrac{D}{\sqrt{3}}=\dfrac{\sqrt{\sqrt{3}+1}+1-\sqrt{\sqrt{3}+1}+1}{\sqrt{3}+1-1}\)
\(\dfrac{D}{\sqrt{3}}=\dfrac{2}{\sqrt{3}}\)
D=2
Lời giải:
Gọi tổng trên là $A$. Ta có:
\(2A>\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{79}+\sqrt{80}}+\frac{1}{\sqrt{80}+\sqrt{81}}\)
\(2A>\frac{\sqrt{2}-1}{(\sqrt{1}+\sqrt{2})(\sqrt{2}-1)}+\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{2}+\sqrt{3})(\sqrt{3}-\sqrt{2})}+\frac{\sqrt{4}-\sqrt{3}}{(\sqrt{3}+\sqrt{4})(\sqrt{4}-\sqrt{3})}+...+\frac{\sqrt{81}-\sqrt{80}}{(\sqrt{80}+\sqrt{81})(\sqrt{81}-\sqrt{80})}\)
\(2A>(\sqrt{2}-1)+(\sqrt{3}-\sqrt{2})+(\sqrt{4}-\sqrt{3})+....+(\sqrt{81}-\sqrt{80})\)
\(2A>\sqrt{81}-1=8\Rightarrow A>4\)
Ta có đpcm.
