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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+..+\frac{1}{72}=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{8.9}\)
\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-...-\frac{1}{9}\)
\(=\frac{1}{3}-\frac{1}{9}=\frac{2}{9}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`@` `\text {Ans}`
`\downarrow`
`a)`
`13/50 + 9% + 41/100 + 0,24`
`= 0,26 + 0,09 + 0,41 + 0,24`
`= (0,26 + 0,24) + (0,09 + 0,41)`
`= 0,5 + 0,5`
`= 1`
`b)`
`2018 \times 2020 - 1/2017 + 2018 \times 2019`
`= 2018 \times (2020 + 2019) - 1/2017`
`= 2018 \times 4039 - 1/2017`
`= 8150702`
`c)`
`1/2 + 1/6 + 1/12 + 1/20 +1/30 +1/42`
`=`\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+\dfrac{1}{6\times7}\)
`=`\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{6}-\dfrac{1}{7}\)
`=`\(1-\dfrac{1}{7}\)
`= 6/7`
\(a,\dfrac{13}{50}+9\%+\dfrac{41}{100}+0,24\\ 0,26+0,09+0,41+0,24\\ =\left(0,26+0,24\right)+\left(0,09+0,41\right)\\ =0,5+0,5\\ =1\\ b,2018\times2020-\dfrac{1}{2017}+2018\times2019\\ =2018\times\left(2020+2019\right)-\dfrac{1}{2017}\\ =2018\times4039-\dfrac{1}{2017}\\ =3150702-\dfrac{1}{2017}\\ c,\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\\ =1-\dfrac{1}{2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}.........+\dfrac{1}{6}-\dfrac{1}{7}\\ =1-\dfrac{1}{7}\\ =\dfrac{6}{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1/2 +1/(2.3) +1/(3.4) + 1/(4.5) + 1/(5.6) + 1/(6.7)
=(2-1)/2 + (3-2)/(2.3) +(4-3)/(3.4) + (5-4)/(4.5) + (6-5)/(5.6) + (7-6)/(6.7)
=1-1/2 +1/2-1/3+....+1/6-1/7
=1-1/7
=6/7
đặt biểu thức trên là A
\(\Rightarrow A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{6.7}\)
\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+....+\frac{7-6}{6.7}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-....+\frac{1}{6}-\frac{1}{7}\)
\(=1-\frac{1}{7}=\frac{6}{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{5}+\frac{2}{3}=\frac{8}{15}\)
\(\frac{3}{4}\div\frac{1}{5}=\frac{15}{4}\)
\(\frac{1}{5}+\frac{2}{3}\div\frac{1}{5}=\frac{53}{15}\)
\(\frac{11}{4}-\frac{2}{3}\div\frac{1}{2}=\frac{17}{12}\)
Nếu mình đúng thì các bạn k mình nhé
\(\frac{1}{5}+\frac{2}{3}=\frac{1.3}{5.3}+\frac{2.5}{3.5}\)\(=\frac{3}{15}+\frac{10}{15}=\frac{13}{15}\)
\(\frac{3}{4}:\frac{1}{5}=\frac{3}{4}.5=\frac{15}{4}\)
\(\frac{1}{5}+\frac{2}{3}:\frac{1}{5}=\frac{1}{5}+\frac{10}{3}=\frac{3}{15}+\frac{50}{15}=\frac{53}{15}\)
\(\frac{11}{4}-\frac{2}{3}:\frac{1}{2}=\frac{11}{4}-\frac{4}{3}=\frac{33}{12}-\frac{16}{12}=\frac{17}{12}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
có thể đây là bài lớp 4 nhưng mình nghĩ là các bạn lớp 5 cũng sẽ khó khăn đó
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)
\(\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{7}=\frac{6}{7}\)
1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42
= 1/1x2 + 1/2x3 + 1/3x4 + 1/4x5 + 1/5x6 + 1/6x7
= 1/1 + 1/7
= 8/7
![](https://rs.olm.vn/images/avt/0.png?1311)
Mình tính từng cái ra nha, từng cái sẽ ra được kết quả của phép tính:
\(1-\dfrac{1}{5}-\dfrac{1}{6}\)
\(=\left(1-\dfrac{1}{5}\right)-\dfrac{1}{6}\)
\(=\left(\dfrac{5}{5}-\dfrac{1}{5}\right)-\dfrac{1}{6}\)
\(=\dfrac{4}{5}-\dfrac{1}{6}\)
\(=\dfrac{24}{30}-\dfrac{5}{30}\)
\(=\dfrac{19}{30}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
đề = \(\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{50.51}\)( áp dụng c.thức tính tổng )
= ..........
= 2 .( \(\frac{1}{2}-\frac{1}{51}\))
= dễ
= 1/2x3 + 1/3x4 + 1/4x5 +...+ 1/9x10 + 1/10x11
= 1/2 - 1/3 + 1/3 - 1/4 +1/4 - 1/5+...+ 1/9 - 1/10 + 1/10 - 1/11
=1/2-1/11 = 9/22
( hình như đề sai bạn ơi)
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