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\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{9}\right)\)
\(=\frac{1}{2}.\frac{8}{9}\)
\(=\frac{4}{9}\)
Đặt: A=1/1.3+1/3.5+1/5.7+1/7.9
2A=2/1.3+2/3.5+2/5.7+2/7.9
2A=1-1/3+1/3-1/5+1/5-1/7+1/7-1/9
2A=1-1/9
2A=8/9
A=4/9
\(\frac{2}{3.5}+\frac{2}{5.7}+........+\frac{2}{37.39}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+......+\frac{1}{37}-\frac{1}{39}\)
\(=\frac{1}{3}-\frac{1}{39}\)
\(=\frac{13}{39}-\frac{1}{39}\)
\(=\frac{12}{39}=\frac{4}{13}\)
ta có A=1/3-1/5+1/5-1/7+1/7-1/9+....+1/37-1/39
=1/3-1/39
=12/39
\(\frac{1}{n\left(n+1\right)}=\frac{n+1-n}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n-1}\)
Áp dụng ta có:
\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
Tính C tương tự, áp dụng:
\(\frac{2}{n\left(n+2\right)}=\frac{n+2-n}{n\left(n+2\right)}=\frac{1}{n}-\frac{1}{n+2}\)
B = 9899/9900
C=I don't know !!
Ủng hộ nhé !
\(8\frac{7}{10}+2\frac{3}{4}=\frac{87}{10}+\frac{11}{4}=\frac{174}{20}+\frac{55}{20}=\frac{229}{20}\)
Bạn chỉ cần đưa về phân số xong tính bình thường. Muốn đổi từ hỗn số sang phân số, ta chỉ cần lấy phần nguyên nhân cho mẫu rồi cộng với tử là xong. Chứ bạn cứ hỏi mấy bài dễ như thế này thì k giỏi đc đâu!!!
b) 2/3.5+2/5/7+2/7.9+2/9.11+...+2/13.15
=1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+..+1/13-1/15
=1/3-1/15
=4/15
c) 2/1.2+2/2.3+2/3.4+..+2/8.9+2/9.10
=2(1/1.2+1/2.3+1/3.4+..+1/8.9+1/9.10)
=2(1-1/2+1/2-1/3+1/3-1/4+..+1/8-1/9+1/9-1/10)
=2(1-1/10)
=2.9/10=9/5
d) 1/3+1/9+1/27+...+1/729
đặt A=1/3^1+1/3^2+1/3^3+..+1/3^6
3A=1+1/3+1/3^2+...+1/3^5
3A-A=1+1/3+1/3^2+...+1/3^5-1/3-1/3^2-1/3^3-...-1/3^6
2A=1-1/3^6
2A=728/729
A=728/729:2
A=364/729
\(\frac{23}{12}\)
\(\frac{314}{105}\)
\(\frac{59}{60}\)
\(\frac{199}{90}\)
\(\frac{1}{18}\)
\(\frac{13}{36}\)
\(\frac{4}{221}\)
\(\frac{4}{85}\)
a)\(=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)
\(=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{2015}\right)\)
\(=\frac{3}{2}\cdot\frac{402}{2015}\)
\(=\frac{603}{2015}\)
b)\(=\frac{4}{5}\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+...+\frac{1}{93}-\frac{1}{98}\right)\)
\(=\frac{4}{5}\left(\frac{1}{3}-\frac{1}{98}\right)\)
\(=\frac{4}{5}\cdot\frac{95}{294}\)
\(=\frac{38}{147}\)
a) Gọi tổng trên là A
A = \(\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}+...+\frac{3}{2013.2015}\)
A == \(\frac{3}{5}-\frac{3}{7}+\frac{3}{7}-\frac{3}{9}+\frac{3}{9}-\frac{3}{11}+...+\frac{3}{2013}-\frac{3}{2015}\)
Vì một số trừ cho a rồi cộng cho a sẽ bằng chính số đó nên:
A = \(\frac{3}{5}-\frac{3}{2015}\)
A = \(\frac{1209}{2015}-\frac{3}{2015}\)
A = \(\frac{1206}{2015}\)
b) Gọi tổng trên là B
B = \(\frac{4}{3.8}+\frac{4}{8.13}+\frac{4}{13.15}+...+\frac{4}{93.98}\)
B = \(\frac{4}{3}-\frac{4}{8}+\frac{4}{8}-\frac{4}{13}+\frac{4}{13}-\frac{4}{15}+...+\frac{4}{93}-\frac{4}{98}\)
Vì một số trừ cho a rồi cộng cho a sẽ bằng chính số đó nên:
B = \(\frac{4}{3}-\frac{4}{98}\)
B = \(\frac{686}{294}-\frac{12}{294}\)
B = \(\frac{674}{294}=\frac{337}{147}\)
\(A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}+\frac{2}{15.17}\)
\(A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{17}\)
\(A=1-\frac{1}{17}\)
\(A=\frac{16}{17}\)
\(B=\frac{4}{1.3}+\frac{4}{3.5}+...+\frac{4}{9.11}+\frac{4}{11.13}\)
\(B=\frac{4}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(B=\frac{4}{2}\left(1-\frac{1}{13}\right)\)
\(B=\frac{4}{2}\cdot\frac{12}{13}\)
\(B=\frac{24}{13}\)
=> A= \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}\)
=> A= \(\frac{1}{1}-\frac{1}{17}\)
=> A= \(\frac{16}{17}\)
\(\Rightarrow B=2.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(\Rightarrow B=2.\left(\frac{1}{1}-\frac{1}{13}\right)\)
\(\Rightarrow B=2.\frac{12}{13}\)
\(\Rightarrow B=\frac{24}{13}\)
A = \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}\)
= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\)
= \(1-\frac{1}{15}\)= \(\frac{14}{15}\)
B = \(\frac{4}{1.3}+\frac{4}{3.5}+...+\frac{4}{9.11}+\frac{4}{11.13}\)
= \(2.\left(\frac{2}{1.3}-\frac{2}{3.5}+...+\frac{2}{11.13}\right)\)
= \(2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{13}\right)\)
= \(2.\left(1-\frac{1}{13}\right)=2.\frac{12}{13}\)
= \(\frac{24}{13}\)
Theo đề bài ta có :
\(A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{13.15}+\frac{2}{15.17}\)
\(A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}\)
\(A=1-\frac{1}{17}\)
\(A=\frac{16}{17}\)
\(B=\frac{4}{1.3}+\frac{4}{3.5}+...+\frac{4}{9.11}+\frac{4}{11.13}\)
\(B=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{13}\right)\)
\(B=\frac{1}{2}\left(1-\frac{1}{3}\right)\)
\(B=\frac{1}{2}.\frac{2}{3}\)
\(B=\frac{1}{3}\)
2+3.52+5.72+...+13.152+15.172
\(A = 1 - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + . . . . + \frac{1}{17}\)
\(A = 1 - \frac{1}{17}\)
\(A = \frac{16}{17}\)
\(B = \frac{4}{1.3} + \frac{4}{3.5} + . . . + \frac{4}{9.11} + \frac{4}{11.13}\)
\(B = \frac{4}{2} \left(\right. 1 - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + . . . + \frac{1}{9} - \frac{1}{11} + \frac{1}{11} - \frac{1}{13} \left.\right)\)
\(B = \frac{4}{2} \left(\right. 1 - \frac{1}{13} \left.\right)\)
\(B = \frac{4}{2} \cdot \frac{12}{13}\)
\(B = \frac{24}{13}\)