\(A=\dfrac{2}{1x3}+\dfrac{2}{3x5}+\dfrac{2}{5x7}+...+\dfrac{2}{21x23}\)

\(A=2x\left(\dfrac{1}{1x3}+\dfrac{1}{3x5}+\dfrac{1}{5x7}+...+\dfrac{1}{21x23}\right)\)

\(A=2x\dfrac{1}{2}x\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{21}-\dfrac{1}{23}\right)\)

\(A=1-\dfrac{1}{23}\)

\(A=\dfrac{22}{23}\)

\(B=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\)

\(B=\dfrac{1}{2x3}+\dfrac{1}{3x4}+\dfrac{1}{4x5}+\dfrac{1}{5x6}+\dfrac{1}{6x7}+\dfrac{1}{7x8}+\dfrac{1}{8x9}+\dfrac{1}{9x10}\)

\(B=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)

\(B=\dfrac{1}{2}-\dfrac{1}{10}\)

\(B=\dfrac{5}{10}-\dfrac{1}{10}\)

\(B=\dfrac{4}{10}\)

\(B=\dfrac{2}{5}\)

\(C=\dfrac{1}{28}+\dfrac{1}{70}+\dfrac{1}{130}+\dfrac{1}{208}\)

\(C=\dfrac{1}{4x7}+\dfrac{1}{7x10}+\dfrac{1}{10x13}+\dfrac{1}{13x16}\)

\(C=\dfrac{1}{3}x\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}\right)\)

\(C=\dfrac{1}{3}x\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\)

\(C=\dfrac{1}{3}x\left(\dfrac{4}{16}-\dfrac{1}{16}\right)\)

\(C=\dfrac{1}{3}x\dfrac{3}{16}\)

\(C=\dfrac{1}{16}\)

mn ơi giúp mình với

Híc híc

a) 7x - 2x = 6¹⁷ : 6¹⁵ + 44

=> 5x = 36 + 44

=> 5x = 80

=> x = 80 : 5 = 16

b) 9^{x - 1} = 18 + 1/9 - 1/9 - 9

=> 9^{x - 1} = 9

=> x - 1 = 1

=> x = 1 + 1 = 2

c) [(6x - 39) : 7] . 4 = 12

=> (6x - 39) : 7 = 12 : 4

=> (6x - 39) : 7 = 3

=> 6x - 39 = 3.7

=> 6x - 39 = 21

=> 6x = 21 + 39

=> 6x = 60

=> x = 60 : 6

=> x = 10

d) 2 - (x - 1) - 3x = 20

=> 2 - x + 1 - 3x = 20

=> 3 - 4x = 20

=> 4x = 3 - 20

=> 4x = -17

=> x = -17 : 4 = -17/4

e) 2|x - 3| + 7 = 5⁶ : 5²

=> 2|x - 3| + 7 = 625

=> 2|x - 3| = 625 - 7

=> 2|x - 3| = 618

=> |x - 3| = 618 : 2

=> |x - 3| = 309

=> \(\orbr{\begin{cases}x-3=309\\x-3=-309\end{cases}}\)

=> \(\orbr{\begin{cases}x=312\\x=-306\end{cases}}\)

Bài này bạn chỉ cần bỏ chút thời gian là làm đc, làm như những bài tìm x bình thường thôi

a) \(\dfrac{13}{20}+\dfrac{3}{5}+x=\dfrac{5}{6}\)

\(\Rightarrow\dfrac{5}{4}+x=\dfrac{5}{6}\)

\(\Rightarrow x=\dfrac{5}{6}-\dfrac{5}{4}\)

\(\Rightarrow x=\dfrac{-5}{12}\)

b) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\dfrac{-1}{3}\)

\(\Rightarrow x+\dfrac{1}{3}=\dfrac{11}{15}\)

\(\Rightarrow x=\dfrac{11}{15}-\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{2}{5}\)

c)\(\dfrac{-5}{8}-x=\dfrac{-3}{20}-\dfrac{-1}{6}\)

\(\dfrac{-5}{8}-x=\dfrac{1}{60}\)

\(\Rightarrow x=\dfrac{-5}{8}-\dfrac{1}{60}\)

\(\Rightarrow x=\dfrac{-77}{120}\)

d) \(\dfrac{3}{5}-x=\dfrac{1}{4}+\dfrac{7}{10}\)

\(\Rightarrow\dfrac{3}{5}-x=\dfrac{19}{20}\)

\(\Rightarrow x=\dfrac{3}{5}-\dfrac{19}{20}\)

\(\Rightarrow x=\dfrac{-7}{20}\)

e) \(\dfrac{-3}{7}-x=\dfrac{4}{5}+\dfrac{-2}{3}\)

\(\Rightarrow\dfrac{-3}{7}-x=\dfrac{2}{15}\)

\(\Rightarrow x=\dfrac{-3}{7}-\dfrac{2}{15}\)

\(\Rightarrow x=\dfrac{-59}{105}\)

g) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)

\(\Rightarrow\dfrac{-5}{6}-x=\dfrac{1}{4}\)

\(\Rightarrow x=\dfrac{-5}{6}-\dfrac{1}{4}\)

\(\Rightarrow x=\dfrac{-13}{12}\)

dễ mà 

\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\)

\(=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)

\(=1-\dfrac{1}{10}\)

\(=\dfrac{10}{10}-\dfrac{1}{10}=\dfrac{9}{10}\)

a, \(\dfrac{1}{2}\) - ( - \(\dfrac{1}{3}\) ) + \(\dfrac{1}{23}\) + \(\dfrac{1}{6}\)

 =  \(\dfrac{5}{6}\)  + \(\dfrac{1}{23}\) + \(\dfrac{1}{6}\)

= 1 + \(\dfrac{1}{23}\)

 = \(\dfrac{24}{23}\) 

b, \(\dfrac{11}{24}\) - \(\dfrac{5}{41}\) + \(\dfrac{13}{24}\) + 0,5 - \(\dfrac{36}{41}\)

= (\(\dfrac{11}{24}\) + \(\dfrac{13}{24}\)) - ( \(\dfrac{5}{41}\) + \(\dfrac{36}{41}\)) + 0,5

= 1 - 1 + 0,5

= 0,5 

c,\(-\dfrac{1}{12}-\left(\dfrac{1}{6}-\dfrac{1}{4}\right)\)

=\(-\dfrac{1}{12}-\left(-\dfrac{1}{12}\right)\)

=0

d, \(\dfrac{1}{6}-\left[\dfrac{1}{6}-\left(\dfrac{1}{4}+\dfrac{9}{12}\right)\right]\)

= \(\dfrac{1}{6}-\left[\dfrac{1}{6}-1\right]\)

= \(\dfrac{1}{6}-\left(-\dfrac{5}{6}\right)\)

= 1

e, 0,5 + \(\dfrac{1}{3}\) + 0,4 + \(\dfrac{5}{7}\) + \(\dfrac{1}{6}\) - \(\dfrac{4}{35}\)

 =    (0,5 + 0,4) + ( \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\)) + (\(\dfrac{5}{7}\)  - \(\dfrac{4}{35}\))

 = 0,9 + \(\dfrac{1}{2}\) + \(\dfrac{3}{5}\)

= 0,9 + 0,5 + 0,6

= 2

Biểu thức = \(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)

= \(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+..+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)

= \(\dfrac{1}{2}-\dfrac{1}{10}=\dfrac{2}{5}\)

\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{110}\)

\(=\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+...+\dfrac{1}{10\times11}\)

= 1 - \(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{10}-\dfrac{1}{11}\)

= \(1-\dfrac{1}{11}=\dfrac{10}{11}\)

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{10.11}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{10}-\dfrac{1}{11}=1-\dfrac{1}{11}=\dfrac{10}{11}\)

Ta có \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};...;\dfrac{1}{20^2}< \dfrac{1}{19.20}\)

Cộng vế với vế ta được 

\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{20^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{19}-\dfrac{1}{20}\)

\(\Rightarrow T< 2-\dfrac{1}{20}=\dfrac{39}{20}\)

mà 39/20 < 8/7 => T < 8/7