Ta có:\(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{200}>\dfrac{1}{200}+\dfrac{1}{200}+...+\dfrac{1}{200}=\dfrac{100}{200}=\dfrac{1}{2}\)

Lại có:

\(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{200}< \dfrac{1}{101}+\dfrac{1}{101}+...+\dfrac{1}{101}=\dfrac{100}{101}\)

Vậy ...

Những dãy trên đều có 100 số hạng.

Ta có:  \(\dfrac{1}{101}>\dfrac{1}{200}\)

Tương tự ta có: \(\dfrac{1}{102}>\dfrac{1}{200}\) ;....; \(\dfrac{1}{199}>\dfrac{1}{200}\)

\(\Rightarrow\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{199}+\dfrac{1}{200}>\dfrac{1}{200}.100\)

\(\Leftrightarrow\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{199}+\dfrac{1}{200}>\dfrac{100}{200}\)

\(\Leftrightarrow\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{199}+\dfrac{1}{200}>\dfrac{1}{2}\left(đpcm\right)\)

\(C=\left(\dfrac{1}{200^2}-1\right)\left(\dfrac{1}{199^2-1}\right)...\left(\dfrac{1}{101^2-1}\right)\)

\(C=\dfrac{1-200^2}{200^2}.\dfrac{1-199^2}{199^2}.\dfrac{1-198^2}{198^2}...\dfrac{1-101^2}{101^2}\)

\(C=\dfrac{\left(1-200\right)\left(1+200\right)}{200^2}.\dfrac{\left(1-199\right)\left(1+199\right)}{199^2}...\dfrac{\left(1-100\right)\left(1+100\right)}{100^2}.\dfrac{\left(1-101\right)\left(1+101\right)}{101^2}\) \(C=\dfrac{-199.201}{200.200}.\dfrac{-198.200}{199.199}.\dfrac{-197.199}{198.198}...\dfrac{-99.101}{100.100}.\dfrac{-100.102}{101.101}\)

\(C=\dfrac{199.201}{200.200}.\dfrac{198.200}{199.199}.\dfrac{197.199}{198.198}...\dfrac{99.101}{100.100}.\dfrac{100.102}{101.101}\)

\(\Rightarrow C=\dfrac{200}{2.101}=\dfrac{201}{202}\)

Câu 2 mik chịu r sorry:(

\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{199}-\dfrac{1}{200}\)

\(=\left(1+\dfrac{1}{3}+...+\dfrac{1}{199}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+..+\dfrac{1}{200}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{199}+\dfrac{1}{200}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{200}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{199}+\dfrac{1}{200}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{100}\right)\)

\(=\dfrac{1}{101}+...+\dfrac{1}{199}+\dfrac{1}{200}\)

Lời giải:

Gọi phân số vế trái là $A$. Gọi tử số là $T$. Xét mẫu số:
\(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)

\(=1-\frac{1}{2}+1-\frac{1}{3}+1-\frac{1}{4}+....+1-\frac{1}{100}\)

\(=99-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(=100-(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100})\)

\(=\frac{1}{2}\left[200-(3+\frac{2}{3}+\frac{2}{4}+...+\frac{2}{100})\right]=\frac{1}{2}T\)

$\Rightarrow A=\frac{T}{\frac{1}{2}T}=2$ 

Ta có đpcm.

Giải:

Vì \(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}=2\) nên phần tử gấp 2 lần phần mẫu

Ta có:

\(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\)  

\(=\dfrac{2.\left[100-\left(\dfrac{3}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{100}\right)\right]}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\)

\(=\dfrac{2.\left[\left(2-\dfrac{3}{2}\right)+\left(1-\dfrac{1}{3}\right)+\left(1-\dfrac{1}{4}\right)+\left(1-\dfrac{1}{5}\right)+...+\left(1-\dfrac{1}{100}\right)\right]}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\) 

\(=\dfrac{2.\left(\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+\dfrac{4}{5}+...+\dfrac{99}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\) 

\(=2\) 

Vậy \(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}=2\left(đpcm\right)\) 

Chúc bạn học tốt!

\(\dfrac{100+\dfrac{99}{2}+\dfrac{98}{3}+...+\dfrac{1}{100}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{101}}-2\)

\(=\dfrac{\left(\dfrac{99}{2}+1\right)+\left(\dfrac{98}{3}+1\right)+...+\left(\dfrac{1}{100}+1\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{101}}-2\)

\(=\dfrac{\dfrac{101}{2}+\dfrac{101}{3}+...+\dfrac{101}{100}+\dfrac{101}{101}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{101}}-2\)

\(=\dfrac{101\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}+\dfrac{1}{101}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{101}}-2\)

\(=101-2\)

\(=99\)