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a) \(A=\dfrac{1}{3}-\dfrac{3}{4}-\left(-\dfrac{3}{5}\right)+\dfrac{1}{72}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\)
\(=\dfrac{1}{3}-\dfrac{3}{4}+\dfrac{3}{5}+\dfrac{1}{72}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\)
\(=\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)-\left(\dfrac{3}{4}+\dfrac{2}{9}+\dfrac{1}{36}\right)+\dfrac{1}{72}\)
\(=\left(\dfrac{5}{15}+\dfrac{9}{15}+\dfrac{1}{15}\right)-\left(\dfrac{27}{36}+\dfrac{8}{36}+\dfrac{1}{36}\right)+\dfrac{1}{72}\)
\(=1-1+\dfrac{1}{72}\)
\(=0+\dfrac{1}{72}=\dfrac{1}{72}\)
b) \(B=\dfrac{1}{5}-\dfrac{3}{7}+\dfrac{5}{9}-\dfrac{2}{9}+\dfrac{7}{13}-\dfrac{2}{11}-\dfrac{5}{9}+\dfrac{3}{7}-\dfrac{1}{5}\)
\(=\left(\dfrac{1}{5}-\dfrac{1}{5}\right)+\left(-\dfrac{3}{7}+\dfrac{3}{7}\right)+\left(\dfrac{5}{9}-\dfrac{5}{9}\right)-\left(\dfrac{2}{9}-\dfrac{7}{13}+\dfrac{2}{11}\right)\)
\(=0+0+0-\left(\dfrac{286}{1287}-\dfrac{693}{1287}+\dfrac{234}{1287}\right)\)
\(=-\left(-\dfrac{173}{1287}\right)\)
\(=\dfrac{173}{1287}\)
c) \(C=\dfrac{1}{100}-\dfrac{1}{100.99}-\dfrac{1}{99.98}-.....-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
\(=\dfrac{1}{100}-\left(\dfrac{1}{100.99}+\dfrac{1}{99.98}+\dfrac{1}{98.97}+...+\dfrac{1}{3.2}+\dfrac{1}{2.1}\right)\)
\(=\dfrac{1}{100}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{97.98}+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\)
\(=\dfrac{1}{100}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=\dfrac{1}{100}-\left(1-\dfrac{1}{100}\right)\)
\(=\dfrac{-49}{50}\)
1. Tính:
a. \(\dfrac{\text{−1 }}{\text{4 }}+\dfrac{\text{5 }}{\text{6 }}=\dfrac{-3}{12}+\dfrac{10}{12}=\dfrac{7}{12}\)
b. \(\dfrac{\text{5 }}{\text{12 }}+\dfrac{\text{-7 }}{8}=\dfrac{10}{24}+\dfrac{-21}{24}=\dfrac{-11}{24}\)
c. \(\dfrac{-7}{6}+\dfrac{-3}{10}=\dfrac{-35}{30}+\dfrac{-9}{30}=\dfrac{-44}{30}=\dfrac{-22}{15}\)
d.\(\dfrac{-3}{7}+\dfrac{5}{6}=\dfrac{-18}{42}+\dfrac{35}{42}=\dfrac{17}{42}\)
2. Tính :
a. \(\dfrac{2}{14}-\dfrac{5}{2}=\dfrac{2}{14}-\dfrac{35}{14}=\dfrac{-33}{14}\)
b.\(\dfrac{-13}{12}-\dfrac{5}{18}=\dfrac{-39}{36}-\dfrac{10}{36}=\dfrac{49}{36}\)
c.\(\dfrac{-2}{5}-\dfrac{-3}{11}=\dfrac{-2}{5}+\dfrac{3}{11}=\dfrac{-22}{55}+\dfrac{15}{55}=\dfrac{-7}{55}\)
d. \(0,6--1\dfrac{2}{3}=\dfrac{6}{10}--\dfrac{5}{3}=\dfrac{3}{5}+\dfrac{5}{3}=\dfrac{9}{15}+\dfrac{25}{15}=\dfrac{34}{15}\)
3. Tính :
a.\(\dfrac{-1}{39}+\dfrac{-1}{52}=\dfrac{-4}{156}+\dfrac{-3}{156}=\dfrac{-7}{156}\)
b.\(\dfrac{-6}{9}-\dfrac{12}{16}=\dfrac{2}{3}-\dfrac{3}{4}=\dfrac{8}{12}-\dfrac{9}{12}=\dfrac{-17}{12}\)
c. \(\dfrac{-3}{7}-\dfrac{-2}{11}=\dfrac{-3}{7}+\dfrac{2}{11}=\dfrac{-33}{77}+\dfrac{14}{77}=\dfrac{-19}{77}\)
d.\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(=\dfrac{1}{1}+\dfrac{1}{10}\)
\(=\dfrac{10}{10}-\dfrac{1}{10}\)
= \(\dfrac{9}{10}\)
Chế Kazuto Kirikaya thử tham khảo thử đi !!!
Mấy câu trên kia dễ rồi mình chữa mình câu \(c\) bài \(3\) thôi nhé Kazuto Kirikaya
d) \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=1-\dfrac{1}{10}\)
\(=\dfrac{9}{10}\)
a) \(\dfrac{-2}{3}+\dfrac{3}{4}-\dfrac{-1}{6}+\dfrac{-2}{5}=\dfrac{1}{12}-\dfrac{-1}{6}+\dfrac{-2}{5}=\dfrac{1}{4}+\dfrac{-2}{5}=\dfrac{-3}{20}\)
b) \(\dfrac{-2}{3}+\dfrac{-1}{5}+\dfrac{3}{4}-\dfrac{5}{6}-\dfrac{-7}{10}=\left(\dfrac{-2}{3}-\dfrac{5}{6}\right)+\left(\dfrac{-1}{5}-\dfrac{-7}{10}\right)+\dfrac{3}{4}\)
\(=\dfrac{-3}{2}+\dfrac{1}{2}-\dfrac{3}{4}\)
= \(=-1-\dfrac{3}{4}\)
\(=\dfrac{-1}{4}\)
c)\(\dfrac{1}{2}-\dfrac{-2}{5}+\dfrac{1}{3}+\dfrac{5}{7}-\dfrac{-1}{6}+\dfrac{-4}{35}+\dfrac{1}{41}\)
= \(\left(\dfrac{1}{2}-\dfrac{-1}{6}+\dfrac{1}{3}\right)+\left(\dfrac{-4}{35}+\dfrac{5}{7}-\dfrac{-2}{5}\right)+\dfrac{1}{41}\)
= \(1+1+\dfrac{1}{41}\)
= \(\dfrac{83}{41}\)
d)\(\dfrac{1}{100.99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
= \(\dfrac{1}{100}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{98}+...+\dfrac{1}{3}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{1}\)
= \(\dfrac{1}{100}-\dfrac{1}{1}\)
= \(\dfrac{-99}{100}\)
d đảo 1/1.2.1/2.3 ... 1/99.1000
=1/1 -1/2 +1/2-1/3 ... -1/99 - 1/1000
=1/1 -1/1000
=999/1000
b) \(B=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2018}\right)\)
\(=\frac{2-1}{2}.\frac{3-1}{3}.\frac{4-1}{4}....\frac{2018-1}{2018}\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2017}{2018}=\frac{1.2.3...2017}{2.3.4...2018}=\frac{1}{2018}\)
c) Giữa các biểu thức là dấu nhân hay dấu cộng vậy bạn?
d)
\(D=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(D=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{100-99}{99.100}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
e) \(E=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{97.99}\)
\(2E=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(2E=\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+....+\frac{99-97}{97.99}\)
\(2E=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
\(\Rightarrow E=\frac{16}{99}\)
Ta có:
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=1-\dfrac{1}{50}\)
\(\Rightarrow A=\dfrac{49}{50}\)
Vậy \(A=\dfrac{49}{50}.\)
a,\(\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{11}{13}+\dfrac{13}{15}+\dfrac{11}{13}-\dfrac{9}{11}+\dfrac{7}{9}-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{1}{3}\)
\(=\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(-\dfrac{3}{5}+\dfrac{3}{5}\right)+.....+\left(-\dfrac{11}{13}+\dfrac{11}{13}\right)+\dfrac{13}{15}\)
\(=0+0+...0+0+\dfrac{13}{15}=\dfrac{13}{15}\)
câu b và c xem lại đề nha
Chúc bạn học tốt!!!
Bài 1:
a: \(=17+\dfrac{2}{31}-\dfrac{15}{17}-6-\dfrac{2}{31}=11-\dfrac{15}{17}=\dfrac{172}{17}\)
b: \(=31+\dfrac{6}{13}+5+\dfrac{9}{41}-36-\dfrac{9}{41}-36-\dfrac{6}{13}\)
=36
c: \(=27+\dfrac{51}{59}-7-\dfrac{51}{59}+\dfrac{1}{3}=20+\dfrac{1}{3}=\dfrac{61}{3}\)
Lời giải:
Đặt \(A=\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-....-\frac{1}{3.2}-\frac{1}{2.1}\)
\(\Rightarrow A+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}=\frac{1}{99.100}\)
\(\Leftrightarrow A+\frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{98-97}{97.98}+\frac{99-98}{98.99}=\frac{1}{99.100}\)
\(\Leftrightarrow A+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}=\frac{1}{99.100}\)
\(\Leftrightarrow A+1-\frac{1}{98}=\frac{1}{99.100}\Rightarrow A=\frac{1}{9900}-\frac{97}{98}\)
\(a)3\dfrac{1}{2}.\dfrac{4}{49}-\left[2,\left(4\right):2\dfrac{5}{11}\right]:\left(\dfrac{-42}{5}\right)\)
\(=\dfrac{7}{2}.\dfrac{4}{49}-\dfrac{88}{27}:\left(\dfrac{-42}{7}\right)\)
\(=\dfrac{2}{7}-\dfrac{-220}{567}\)
\(=\dfrac{382}{567}\)
các phần con lại dễ nên bn tự lm đi nhé mk bn lắm
Chúc bạn học tốt!
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.....+\dfrac{1}{9.10}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.....+\dfrac{1}{9}-\dfrac{1}{10}\)
\(A=1-\dfrac{1}{10}=\dfrac{9}{10}\)
\(B=\dfrac{1}{100}-\dfrac{1}{100.99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-.....-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
\(B=\dfrac{1}{100}-\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{99}+\dfrac{1}{98}-\dfrac{1}{98}+\dfrac{1}{97}-.....+\dfrac{1}{3}-\dfrac{1}{2}+\dfrac{1}{2}-1\)\(B=0-1=-1\)
cám ơn bn