S=(1-2)+(3-4)+(5-6)+...+(199-200)

S=(-1)+(-1)+...+(-1)

S=(-1).100=-100

S=1+(2-3)+(-4+5)+...+(98-99)+(-100+101)

S=1+(-1)+1+..+(-1)+1

S=1+25.(-1)+25.1

S=1+(-25)+25

S=1+0

=1

Nhanh nha mk cần gấp đó!

Ta luôn chứng minh được: Nếu \(\frac{a}{b}>1\Leftrightarrow\frac{a}{b}>\frac{a+1}{b+1}\)và \(\frac{a}{b}< \frac{a-1}{b-1}\)

Áp dụng điều trên ta có:

\(S=\frac{2}{1}.\frac{4}{3}.\frac{6}{5}...\frac{200}{199}\)

=> \(S>\frac{3}{2}.\frac{5}{4}.\frac{7}{6}...\frac{201}{200}\)

=> \(S^2>\frac{2}{1}.\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.\frac{6}{5}.\frac{7}{6}...\frac{200}{199}.\frac{201}{200}\)

=> S² > 201 > 200 (1)

\(S=\frac{2}{1}.\frac{4}{3}.\frac{6}{5}...\frac{200}{199}\)

=> \(S< \frac{2}{1}.\frac{3}{2}.\frac{5}{4}...\frac{199}{198}\)

=> \(S^2< \frac{2}{1}.\frac{2}{1}.\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.\frac{6}{5}...\frac{199}{198}.\frac{200}{199}\)

=> \(S^2< 400\)(2)

Từ (1) và (2) => 200 < S² < 400 (đpcm)

anh học trường cấp hai nào thế?

A = \(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{199.200}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)

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

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

\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)

Lại có B = \(\frac{1}{101.200}+\frac{1}{102.199}+...+\frac{1}{200.101}\)

=> 301B = \(\frac{301}{101.200}+\frac{301}{102.199}+...+\frac{301}{200.101}\) 

=> 301B = \(\frac{1}{101}+\frac{1}{200}+\frac{1}{102}+\frac{1}{199}+...+\frac{1}{200}+\frac{1}{101}=2\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\right)\)

=> B = \(\frac{2}{301}\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\right)\)

Khi đó \(\frac{A}{B}=\frac{\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\right)}{\frac{2}{301}\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\right)}=\frac{1}{\frac{2}{301}}=\frac{301}{2}=150,5\)

-200 hả bạn nếu đúng **** cho mình nhé 

\(S^2=\left(\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}\cdot...\cdot\dfrac{199}{200}\right)\left(\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}\cdot...\cdot\dfrac{199}{200}\right)\\ \text{Ta có:}\\ \dfrac{1}{2}< \dfrac{2}{3}\\ \dfrac{3}{4}< \dfrac{4}{5}\\ \dfrac{5}{6}< \dfrac{6}{7}\\ ...\\ \dfrac{199}{200}< \dfrac{200}{201}\\ \Rightarrow S^2< \left(\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}\cdot...\cdot\dfrac{199}{200}\right)\left(\dfrac{2}{3}\cdot\dfrac{4}{5}\cdot\dfrac{6}{7}\cdot...\cdot\dfrac{200}{201}\right)\\ \Leftrightarrow S^2< \dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{199}{200}\cdot\dfrac{200}{201}\\ \Leftrightarrow S^2< \dfrac{1\cdot2\cdot3\cdot...\cdot200}{2\cdot3\cdot4\cdot...\cdot201}\\ \Leftrightarrow S^2< \dfrac{1}{201}< \dfrac{1}{200}\)

Vậy ...

(1-2)+(3-4)+.......+(199-200)

=(-1)+(-1)+.........+(-1)        

(100 số âm 1)

=-100

\(1-2+3-4+5-6+....+199-200\)

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

\(=\left(-1\right)+\left(-1\right)+.....+\left(-1\right)\left(100 số\right)\)

\(=-100\)

A=(1-2)+(3-4)+...+(199-200)=(-1)+(-1)+...+(-1)=(-1)*100=-100

\(S=1-2+3-4+...+199-200+201\)

\(=\left(1-2\right)+\left(3-4\right)+...+\left(199-200\right)+201\)

\(=1+1+...+1+201\)

\(=\dfrac{200}{2}+201\)

\(=301\)