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15 tháng 6 2023

\(a,\dfrac{3}{5}+\dfrac{3}{5\cdot9}+\dfrac{3}{9\cdot13}+....+\dfrac{3}{97\cdot101}\)

\(=\dfrac{3}{4}\cdot\left(\dfrac{4}{5}+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+....+\dfrac{4}{97\cdot101}\right)\)

\(=\dfrac{3}{4}\cdot\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+....+\dfrac{1}{97}-\dfrac{1}{101}\right)\)

\(=\dfrac{3}{4}\cdot\left(1-\dfrac{1}{101}\right)\)

\(=\dfrac{3}{4}\cdot\dfrac{100}{101}\)

\(=\dfrac{75}{101}\)

\(b,\left(1+\dfrac{1}{2}\right)\cdot\left(1+\dfrac{1}{3}\right)\cdot\left(1+\dfrac{1}{4}\right)\cdot....\cdot\left(1+\dfrac{1}{99}\right)\)

\(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot....\cdot\dfrac{100}{99}\)

\(=\dfrac{100}{2}=50\)

15 tháng 6 2023

Tính nhanh:

a) \(\dfrac{3}{5}+\dfrac{3}{5.9}+\dfrac{3}{9.13}+...+\dfrac{3}{97.101}\)

\(\dfrac{3}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{97}-\dfrac{1}{101}\right)\)

\(\dfrac{3}{4}\left(1-\dfrac{1}{101}\right)\)

\(\dfrac{3}{4}\times\dfrac{100}{101}\)

\(\dfrac{75}{101}\)

b) \(\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{4}\right)...\left(\dfrac{1}{98}+1\right)\left(\dfrac{1}{99}+1\right)\)

\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}...\dfrac{99}{98}.\dfrac{100}{99}\)

\(=\dfrac{3.4.5...99.100}{2.3.4...98.99}\)

\(=\dfrac{100}{2}\)

\(=50\)

14 tháng 11 2018

1)A=987

16 tháng 6 2023

\(\left(1-\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{4}\right)\cdot...\cdot\left(1-\dfrac{1}{99}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\cdot...\cdot\dfrac{97}{98}\cdot\dfrac{98}{99}\)

\(=\dfrac{1\cdot2\cdot3\cdot...\cdot98}{2\cdot3\cdot4\cdot...\cdot99}\)

\(=\dfrac{1}{99}\)

16 tháng 6 2023

\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}.....\dfrac{99}{98}.\dfrac{100}{99}=\dfrac{100}{2}=50\)

9 tháng 3 2018

a. \(A=\dfrac{3}{2.5}+\dfrac{3}{5.8}+......+\dfrac{3}{17.20}\)

\(=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+......+\dfrac{1}{17}-\dfrac{1}{20}\)

\(=\dfrac{1}{2}-\dfrac{1}{20}\)

\(=\dfrac{9}{20}\)

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

\(=\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)

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

\(=\dfrac{1}{4}-\dfrac{1}{10}\)

\(=\dfrac{3}{20}\)

c. \(C=\dfrac{4^2}{1.5}+\dfrac{4^2}{5.9}+......+\dfrac{4^2}{45.49}\)

\(=4\left(\dfrac{4}{1.5}+\dfrac{4}{5.9}+....+\dfrac{4}{45.49}\right)\)

\(=4\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+.....+\dfrac{1}{45}-\dfrac{1}{49}\right)\)

\(=4\left(1-\dfrac{1}{49}\right)\)

\(=4.\dfrac{48}{49}\)

\(=\dfrac{192}{49}\)

`@` `\text {Ans}`

`\downarrow`

`a)`

\(\left(\dfrac{7}{8}-\dfrac{3}{4}\right)\cdot1\dfrac{1}{3}-\dfrac{2}{3}\cdot0,5\)

`=`\(\dfrac{1}{8}\cdot\dfrac{4}{3}-\dfrac{1}{3}\)

`=`\(\dfrac{1}{6}-\dfrac{1}{3}=-\dfrac{1}{6}\)

`b)`

\(\left(2+\dfrac{5}{6}\right)\div1\dfrac{1}{5}+\left(-\dfrac{7}{12}\right)\)

`=`\(\dfrac{17}{6}\div1\dfrac{1}{5}-\dfrac{7}{12}\)

`=`\(\dfrac{85}{36}-\dfrac{7}{12}=\dfrac{16}{9}\)

`c)`

\(75\%-1\dfrac{1}{2}+0,5\div\dfrac{5}{12}\)

`=`\(-\dfrac{3}{4}+\dfrac{6}{5}=\dfrac{9}{20}\)

23 tháng 7 2023

a) \(\left(\dfrac{7}{8}-\dfrac{3}{4}\right).1\dfrac{1}{3}-\dfrac{2}{3}.0,5\)

\(=\left(\dfrac{7}{8}-\dfrac{6}{8}\right).\dfrac{4}{3}-\dfrac{2}{3}.\dfrac{1}{2}\)

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

\(=\dfrac{1}{6}-\dfrac{1}{3}\)

\(=\dfrac{-1}{6}\)

b) \(\left(2+\dfrac{5}{6}\right):1\dfrac{1}{5}+\dfrac{-7}{12}\)

\(=\left(\dfrac{12}{6}+\dfrac{5}{6}\right):\dfrac{6}{5}+\dfrac{-7}{12}\)

\(=\dfrac{17}{6}.\dfrac{5}{6}+\dfrac{-7}{12}\)

\(=\dfrac{85}{36}+\dfrac{-7}{12}\)

\(=\dfrac{16}{9}\)

c) \(75\%-1\dfrac{1}{2}+0,5:\dfrac{5}{12}\)

\(=\dfrac{3}{4}-\dfrac{3}{2}+\dfrac{1}{2}.\dfrac{12}{5}\)

\(=\dfrac{3}{4}-\dfrac{6}{4}+\dfrac{6}{5}\)

\(=\dfrac{-3}{4}+\dfrac{6}{5}\)

\(=\dfrac{9}{20}\)

4 tháng 10 2022

ai bt tự làm

 

15 tháng 4 2023

ngu tự chịu

5 tháng 8 2018

\(x-\frac{37}{45}=\frac{4}{5.9}+\frac{4}{9.13}+.....+\frac{4}{41.45}\)

\(\Rightarrow x-\frac{37}{45}=\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\)

\(\Rightarrow x-\frac{37}{45}=\frac{1}{5}-\frac{1}{45}\)

\(\Rightarrow x-\frac{37}{45}=\frac{8}{45}\)

\(\Rightarrow x=\frac{37}{45}+\frac{8}{45}\)

\(\Rightarrow x=1\)

3 tháng 2 2019

Đặt \(B=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\)

\(=\left(1+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+\left(\frac{1}{5}+\frac{1}{95}\right)+...+\left(\frac{1}{49}+\frac{1}{51}\right)\)

\(=\frac{100}{99}+\frac{100}{3\times97}+\frac{100}{5\times95}+...+\frac{100}{49\times51}\)

\(=100\left(\frac{1}{99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{49\times51}\right)\)

Đặt \(C=\frac{1}{1\times99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{97\times3}+\frac{1}{99\times1}\)

\(=2\left(\frac{1}{99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{49\times51}\right)\)

\(A=\frac{B}{6}=\frac{100}{2}=50\)

Vậy \(A=50\)

28 tháng 6 2021

`a)25/(x+1)-1 1/6=-1/3-0,5`

`=>25/(x+1)=-1/3-1/2+1+1/6`

`=>25/(x+1)=1/3`

`=>75=x+1`

`=>x=74`

Vậy `x=74`

`b)(2x+25 3/5)^2-9/25=0`

`=>(2x+128/5)=9/25`

`**2x+128/5=3/5`

`=>2x=-125/5=-25`

`=>x=-25/2`

`**2x+128/5=-3/5`

`=>2x=-131/5`

`=>x=-131/10`

28 tháng 6 2021

cám ơn bạn nhiều lắm ạ vui