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\(A = \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \frac{1}{4} - \frac{1}{5} + ... + \frac{1}{999} - \frac{1}{200}\)

\(A = \frac{1}{2} - \frac{1}{200}\)

\(A = \frac{100}{200} - \frac{1}{200}\)

\(A = \frac{99}{200}\)

24 tháng 3 2022

`Answer:`

Bài 1.

a. \(\frac{5}{3}+\frac{1}{4}.\frac{8}{3}=\frac{5}{3}+\frac{2}{3}=\frac{7}{3}\)

b. \(\frac{5}{7}-\frac{9}{8}:\frac{3}{2}=\frac{5}{7}-\frac{3}{4}=-\frac{1}{28}\)

c. \(\frac{5}{9}.\left(-\frac{7}{13}\right)+\frac{5}{9}.\left(-\frac{9}{13}\right)+\frac{5}{9}.\frac{3}{13}=\frac{5}{9}.\left(-\frac{7}{13}+-\frac{9}{13}+\frac{3}{13}\right)=\frac{5}{9}.-1=-\frac{5}{9}\)

Bài 2.

a. \(x-\frac{1}{2}=\frac{3}{10}.\frac{5}{6}\)

\(\Leftrightarrow x-\frac{1}{2}=\frac{1}{4}\)

\(\Leftrightarrow x=\frac{3}{4}\)

b. \(\frac{x}{5}=-\frac{3}{14}.\frac{7}{3}\)

\(\Leftrightarrow\frac{x}{5}=-\frac{1}{2}\)

\(\Leftrightarrow x=\frac{\left(-1\right).5}{2}\)

\(\Leftrightarrow x=-\frac{5}{2}\)

c. \(x+\frac{2}{3}=\frac{9}{15}.\frac{5}{27}\)

\(\Leftrightarrow x+\frac{2}{3}=\frac{1}{9}\)

\(\Leftrightarrow x=\frac{1}{9}-\frac{2}{3}\)

\(\Leftrightarrow x=-\frac{5}{9}\)

d. \(\frac{5}{9}-x=-\frac{3}{9}\)

\(\Leftrightarrow-x=-\frac{3}{9}-\frac{5}{9}\)

\(\Leftrightarrow-x=-\frac{8}{9}\)

\(\Leftrightarrow x=\frac{8}{9}\)

b. \(x+-\frac{7}{13}=-\frac{21}{13}\)

\(\Leftrightarrow x=-\frac{21}{13}-\frac{-7}{13}\)

\(\Leftrightarrow x=-\frac{14}{13}\)

24 tháng 3 2022

`Answer:`

b. \(-\left(2a-b-2c\right)+\left(-a-b\right)-\left(-2+2b-c\right)+b-1\)

\(=-2a+b+2c+-a-b+2-2b+c+b-1\)

\(=-2a-a+b-b-2b+b+2c+c+2-1\)

\(=-3a-2b+3c+1\)

c. \(\left(-46-c\right)-\left(-2+b-3\right)+\left(2c+b-a+4\right)\)

\(=-46-c+2-b+3+2c+b-a+4\)

\(=-a-b+b-c+2c-46+3+4+2\)

\(=-a+c-37\)

30 tháng 3 2022

a) \(A=\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+......+\frac{1}{2017.2022}\)

\(5A=5.\left(\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+.....+\frac{1}{2017.2022}\right)\)

\(5A=\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+......+\frac{5}{2017.2022}\)

\(5A=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+........+\frac{1}{2017}-\frac{1}{2022}\)

\(5A=1-\frac{1}{2022}\)

\(5A=\frac{2022}{2022}-\frac{1}{2022}\)

\(5A=\frac{2021}{2022}\)

\(A=\frac{2021}{2022}\div5\)

\(A=\frac{20201}{10110}\)

TL: 

\(\frac{5}{6}=\frac{1}{2}+\frac{1}{3}\) 

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HT