Question

[Thương Thương]

Chứng minh rằng:

a) \(\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{8}< 24\)

b) \(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+...+\dfrac{1}{\sqrt{100}}>10\)

c) \(\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}+\sqrt{50}< 30\)

Answer 1

[Trần Quốc Lộc]

\(\text{c) }\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}+\sqrt{50}< 30\)

Ta có : \(6< 6.25\Rightarrow\sqrt{6}< \sqrt{6.25}\Rightarrow\sqrt{6}< 2.5\)

\(12< 12.25\Rightarrow\sqrt{12}< \sqrt{12.25}\Rightarrow\sqrt{12}< 3.5\)

\(20< 20.25\Rightarrow\sqrt{20}< \sqrt{20.25}\Rightarrow\sqrt{20}< 4.5\)

\(30< 30.25\Rightarrow\sqrt{30}< \sqrt{30.25}\Rightarrow\sqrt{30}< 5.5\)

\(42< 42.25\Rightarrow\sqrt{42}< \sqrt{42.25}\Rightarrow\sqrt{42}< 6.5\)

\(50< 56.5\Rightarrow\sqrt{50}< \sqrt{56.25}\Rightarrow\sqrt{50}< 7.5\) \(\left(1\right)\)

Từ \(\left(1\right)\) suy ra :

\(\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}+\sqrt{50}< 2.5+3.5+4.5+5.5+6.5+7.5\)

\(\Rightarrow\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}+\sqrt{50}< 30\) \(\left(ĐPCM\right)\)

Vậy \(\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}+\sqrt{50}< 30\)

Answer 2

[Trần Quốc Lộc]

\(\)\(\text{a) }\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{8}< 24\)

Ta có : \(1< 9\Rightarrow\sqrt{1}< \sqrt{9}\Rightarrow\sqrt{1}< 3\)

\(2< 9\Rightarrow\sqrt{2}< \sqrt{9}\Rightarrow\sqrt{2}< 3\)

\(3< 9\Rightarrow\sqrt{3}< \sqrt{9}\Rightarrow\sqrt{3}< 3\)

\(...\)

\(8< 9\Rightarrow\sqrt{8}< \sqrt{9}\Rightarrow\sqrt{8}< 3\) \(\left(1\right)\)

Từ \(\left(1\right)\) suy ra :

\(\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{8}< 3+3+...+3_{\left(\text{8 số hạng 3}\right)}\) \(\) \(\)

\(\) \(\Rightarrow\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{8}< 3\cdot8\)

\(\Rightarrow\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{8}< 24\) \(\left(ĐPCM\right)\)

Vậy \(\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{8}< 24\)

\(\text{b) }\dfrac{1}{\sqrt{10}}+\dfrac{1}{\sqrt{20}}+...\dfrac{1}{\sqrt{100}}>10\)

Ta có : \(1< 100\Rightarrow\sqrt{1}< \sqrt{100}\Rightarrow\dfrac{1}{\sqrt{1}}< \dfrac{1}{\sqrt{100}}\)

\(2< 100\Rightarrow\sqrt{2}< \sqrt{100}\Rightarrow\dfrac{1}{\sqrt{2}}< \dfrac{1}{\sqrt{100}}\)

\(...\)

\(100=100\Rightarrow\sqrt{100}=\sqrt{100}\dfrac{1}{\sqrt{100}}=\dfrac{1}{\sqrt{100}}\) \(\left(1\right)\)

Từ \(\left(1\right)\) suy ra :

\(\dfrac{1}{\sqrt{10}}+\dfrac{1}{\sqrt{20}}+...\dfrac{1}{\sqrt{100}}>\dfrac{1}{\sqrt{100}}+\dfrac{1}{\sqrt{100}}+...+\dfrac{1}{\sqrt{100}}_{\left(\text{100 số hạng}\dfrac{1}{\sqrt{100}}\right)}\)

\(\Rightarrow\dfrac{1}{\sqrt{10}}+\dfrac{1}{\sqrt{20}}+...\dfrac{1}{\sqrt{100}}>\dfrac{1}{\sqrt{100}}\cdot100\)

\(\Rightarrow\dfrac{1}{\sqrt{10}}+\dfrac{1}{\sqrt{20}}+...\dfrac{1}{\sqrt{100}}>\dfrac{10}{\sqrt{100}}\)

\(\Rightarrow\dfrac{1}{\sqrt{10}}+\dfrac{1}{\sqrt{20}}+...\dfrac{1}{\sqrt{100}}>10\) \(\left(ĐPCM\right)\)

Vậy \(\dfrac{1}{\sqrt{10}}+\dfrac{1}{\sqrt{20}}+...\dfrac{1}{\sqrt{100}}>10\)

\(\)