đề là gì vậy bạn

có ai giải được câu này không?

   \(\frac{1}{1x2}+\frac{1}{3x4}+....+\frac{1}{49x50}\)

\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{49}-\frac{1}{50}\)

\(=\left(1+\frac{1}{3}+...+\frac{1}{49}\right)+\left(-\frac{1}{2}-\frac{1}{4}-.....-\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+......+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}......+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{49}+\frac{1}{50}\right)-\left(1+\frac{1}{2}+......+\frac{1}{25}\right)\)

\(=\frac{1}{26}+\frac{1}{27}+....+\frac{1}{50}\left(đpcm\right)\)

c) 

\(C=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{19.21}\)

   \(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\right)\)

   \(=\frac{1}{2}.\left(1-\frac{1}{21}\right)\)

   \(=\frac{1}{2}.\frac{20}{21}\)

   \(=\frac{10}{21}\)

\(A\)= \(\frac{1}{3.4}+\frac{1}{4.5}+..+\frac{1}{49.50}=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}=\)\(\frac{1}{3}-\frac{1}{50}=\frac{50}{150}-\frac{3}{150}=\frac{47}{150}\)

a)

 \(A=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{49.50}\)

    \(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}\)

    \(=\frac{1}{3}-\frac{1}{50}\)

    \(=\frac{47}{150}\)

b) 

 \(B=\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{19.20}\)

    \(=3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}\right)\)

    \(=3.\left(1-\frac{1}{20}\right)\)

    \(=3.\frac{19}{20}\)

    \(=\frac{57}{20}\)

Lời giải:

$A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}$

$< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{25.26}$

$=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{26-25}{25.26}$

$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{25}-\frac{1}{26}$

$=1-\frac{1}{26}< 1$ (đpcm)

Ta có công thức \(\frac{a}{b.c}=\frac{a}{c-b}.\left(\frac{1}{b}-\frac{1}{c}\right)\)

Dựa vào công thức trên, ta có

\(\frac{1}{1.2}=\frac{1}{2-1}.\left(1-\frac{1}{2}\right)\)

\(\frac{1}{2.3}=\frac{1}{3-2}.\left(\frac{1}{2}-\frac{1}{3}\right)\)

............................................

\(\frac{1}{49.50}=\frac{1}{50-49}.\left(\frac{1}{49}-\frac{1}{50}\right)\)

\(A=1.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}\right)\)

\(\Rightarrow A=1-\frac{1}{50}=\frac{49}{50}\)

chắc chắn bạn ạ, ai thấy đúng hì ủng hộ nha

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}=1-\frac{1}{50}=\frac{49}{50}\)\(\frac{49}{50}\)

A = \(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{49\times50}\)

A = \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

A = \(\frac{1}{1}-\frac{1}{50}\)

A = \(\frac{50}{50}-\frac{1}{50}\)

A = \(\frac{49}{50}\)

Ta có : 

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(A=1-\frac{1}{50}\)

\(A=\frac{49}{50}\)

Vậy \(A=\frac{49}{50}\)

Chúc bạn học tốt ~ 

= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/49 - 1/50

= 1/1 - 1/50

= 49/50
 

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=1-\frac{1}{50}=\frac{49}{50}\)

k mk nha

thanks