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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{5}+\frac{4}{10}+\frac{9}{15}+\frac{16}{20}+\frac{36}{30}+\frac{64}{40}+\frac{81}{45}\)
\(=\frac{33}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{-1}{20}+\dfrac{-1}{30}+\dfrac{-1}{42}+\dfrac{-1}{56}+\dfrac{-1}{72}+\dfrac{-1}{90}\)
\(=\dfrac{-1}{4.5}+\dfrac{-1}{5.6}+\dfrac{-1}{6.7}+\dfrac{-1}{7.8}+\dfrac{-1}{8.9}+\dfrac{-1}{9.10}\)
\(=\dfrac{-1}{4}-\dfrac{-1}{5}+\dfrac{-1}{5}-\dfrac{-1}{6}+\dfrac{-1}{6}-\dfrac{-1}{7}+\dfrac{-1}{7}-\dfrac{-1}{8}+\dfrac{-1}{8}-\dfrac{-1}{9}+\dfrac{-1}{9}-\dfrac{-1}{10}\)
\(=\dfrac{-1}{4}-\dfrac{-1}{10}\)
\(=\dfrac{-3}{20}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1: Ta có: \(20-2\left(x+4\right)=4\)
\(\Leftrightarrow2\left(x+4\right)=16\)
\(\Leftrightarrow x+4=8\)
hay x=4
5: Ta có: \(\left(x+1\right)^3=27\)
\(\Leftrightarrow x+1=3\)
hay x=2
![](https://rs.olm.vn/images/avt/0.png?1311)
A = 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56
A = 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + 1/7.8
A = 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8
A = 1 + ( -1/2 + 1/2 ) + ( -1/3 + 1/3 ) + ( -1/4 + 1/4 ) + ( -1/5 + 1/5 ) + ( -1/6 + 1/6 ) + ( -1/7 + 1/7 ) - 1/8
A = 1 + 0 + 0 + 0 + 0 + 0 + 0 - 1/8
A = 1 - 1/8
A = 7/8
* Sửa đề tí nhé
B = 3/2 - 5/6 + 7/12 - 9/20 + 11/30 - 13/42 + 15/56
B = 3/1.2 - 5/2.3 + 7/3.4 - 9/4.5 + 11/5.6 - 13/6.7 + 15/7.8
B = 3 - 3/2 - 5/2 - ( -5/3 ) + 7/3 - 7/4 - 9/4 - ( -9/5 ) + 11/5 - 11/6 - 13/6 - ( -13/7 ) + 15/7 - 15/8
B = 3 - 3/2 - 5/2 + 5/3 + 7/3 - 7/4 - 9/4 + 9/5 + 11/5 - 11/6 - 13/6 + 13/7 + 15/7 - 15/8
B = 3 + ( -3/2 - 5/2 ) + ( 5/3 + 7/3 ) + ( -7/4 - 9/4 ) + ( 9/5 + 11/5 ) + ( -11/6 - 13/6 ) + ( 13/7 + 15/7 ) - 15/8
B = 3 + -4 + 4 + -4 + 4 + -4 + 4 - 15/8
B = 3 + 0 + 0 + 0 - 15/8
B = 3 - 15/8
B = 9/8
![](https://rs.olm.vn/images/avt/0.png?1311)
1)C= 1/5+1/10+1/20+1/40+...+1/1280
\(=\frac{1}{5}\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\right)\)
Đặt cái trong ngoặc là A ta có:\(2A=2+1+...+\frac{1}{2^7}\)
\(2A-A=\left(2+1+...+\frac{1}{2^7}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2^8}\right)\)
\(A=2-\frac{1}{2^8}\).Thay A vào ta được:\(C=\frac{1}{5}\left(2-\frac{1}{2^8}\right)=\frac{1}{5}\cdot\frac{511}{256}=\frac{511}{1280}\)
2)D= 2/1*3+2/3*5+2/5*10+2/7*9+2/9*11+2/11*18+2/13*15
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{13}-\frac{1}{15}\)
\(=1-\frac{1}{15}\)
\(=\frac{14}{15}\)
3)E= 4/3*7+4/7*11+4/11*15+4/15*19+4/19*23+4/23*27
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}\)
\(=\frac{1}{3}-\frac{1}{27}\)
\(=\frac{8}{27}\)
4)G= 1/2+1/6+1/12+1/20+1/30+1/42+...+1/110
\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\)
\(=1-\frac{1}{11}\)
\(=\frac{10}{11}\)
5)H= 3/1*2+3/2*3+3/3*4+3/4*5+...+3/9*10
\(=3\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=3\left(1-\frac{1}{10}\right)\)
\(=3\times\frac{9}{10}\)
\(=\frac{27}{10}\).Lần sau bạn đăng ít một thôi nhé
\(\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{120}\)
Nhìn qua đề bài thì, ta thấy phân số chưa theo quy luật. Vì vậy nhân phân số với 2 để các phân số có cùng chung quy luật.
\(=\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{240}\)
Sau đó, thấy các phân số có chung số 2 thì bỏ 2 ra ngoài:
\(=2.\left(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+...+\dfrac{1}{240}\right)\)
\(=2.\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{15.16}\right)\)
....
Chúc bạn học tốt
Ta biết : 2(a+b) = 2a + 2b
Tương tự như vậy
\(2\cdot\left(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+....+\dfrac{1}{240}\right)\)
\(=2\cdot\dfrac{1}{20}+2\cdot\dfrac{1}{30}+2\cdot\dfrac{1}{42}+....+2\cdot\dfrac{1}{240}\)
\(=\dfrac{2}{2\cdot10}+\dfrac{2}{2\cdot15}+\dfrac{2}{2\cdot21}+....+\dfrac{2}{2\cdot120}\)
\(=\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+.....+\dfrac{1}{120}\)