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2 tháng 8 2017

\(C=\frac{7^2}{2.9}+\frac{7^2}{9.16}+\frac{7^2}{16.23}+...+\frac{7^2}{65.72}\)

\(C=\frac{7^2}{7}.\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+\frac{1}{16}-\frac{1}{23}+...+\frac{1}{65}-\frac{1}{72}\right)\)

\(C=7.\left(\frac{1}{2}-\frac{1}{72}\right)\)

\(C=7.\frac{35}{72}=\frac{245}{72}\)

2 tháng 8 2017

Ta có : \(C=\frac{7^2}{2.9}+\frac{7^2}{9.16}+\frac{7^2}{16.23}+.....+\frac{7^2}{65.72}\)

\(\Rightarrow C=7\left(\frac{7}{2.9}+\frac{7}{9.16}+\frac{7}{16.23}+.....+\frac{7}{65.72}\right)\)

\(\Rightarrow C=7\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+.....+\frac{1}{65}-\frac{1}{72}\right)\)

\(\Rightarrow C=7\left(\frac{1}{2}-\frac{1}{72}\right)\)

\(\Rightarrow C=7.\frac{35}{72}=\frac{245}{72}\)

18 tháng 7 2017

\(C=\frac{7^2}{2.9}+\frac{7^2}{9.16}+\frac{7^2}{16.23}+....+\frac{7^2}{65.72}\)

\(C=\frac{7^2}{7}\cdot\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+\frac{1}{16}-\frac{1}{23}+....+\frac{1}{65}-\frac{1}{72}\right)\)

\(C=7\cdot\left(\frac{1}{2}-\frac{1}{72}\right)\)

\(C=7\cdot\frac{35}{72}=\frac{245}{72}\)

17 tháng 7 2017

C = 49(1/2.9 ... 1/65.72)

C = 49(1/2 - 1/9 +....+ 1/65 - 1/72)

C = 49( 1/2 - 1/72)

C = bạn tự tính nhé

Có j không hiểu thì Ib mình

9 tháng 12 2017

\(\dfrac{7^2}{2.9}+\dfrac{7^2}{9.16}+\dfrac{7^2}{16.23}+...+\dfrac{7^2}{65.72}\)

\(=7^2\left(\dfrac{1}{2.9}+\dfrac{1}{9.16}+\dfrac{1}{16.23}+...+\dfrac{1}{65.72}\right)\)

\(=7^2\left(\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{23}+...+\dfrac{1}{65}-\dfrac{1}{72}\right)\)

\(=7^2\left(\dfrac{1}{2}-\dfrac{1}{72}\right)\)

\(=49\left(\dfrac{35}{72}\right)\)

\(=\dfrac{1715}{72}\)

9 tháng 12 2017

\(l=\dfrac{7^2}{2.9}+\dfrac{7^2}{9.16}+\dfrac{7^2}{16.23}+...+\dfrac{7^2}{65.72}\)

\(=7\left(\dfrac{7}{2.9}+\dfrac{7}{9.16}+\dfrac{7}{16.23}+...+\dfrac{7}{65.72}\right)\)

\(=7\left(\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{23}+...+\dfrac{1}{65}-\dfrac{1}{72}\right)\)

\(=7\left(\dfrac{1}{2}-\dfrac{1}{72}\right)=7\left(\dfrac{36}{72}-\dfrac{1}{72}\right)=7.\dfrac{35}{72}=\dfrac{245}{72}\)

26 tháng 2 2017

Ta có:

P=\(\frac{1}{3.10}\)+\(\frac{1}{10.17}\)+\(\frac{1}{17.24}\)+......+\(\frac{1}{73.80}\)-\(\frac{1}{2.9}\)-\(\frac{1}{9.16}\)-\(\frac{1}{16.23}\)-\(\frac{1}{23.30}\))

P=\(\frac{1}{7}\)\(\times\)(\(\frac{7}{3.10}\)+\(\frac{7}{10.17}\)+\(\frac{7}{17.24}\)+......\(\frac{7}{73.80}\)-\(\frac{7}{2.9}\)-\(\frac{7}{9.16}\)-\(\frac{7}{16.23}\)-\(\frac{7}{23.30}\))

P=\(\frac{1}{7}\)\(\times\)(\(\frac{1}{3}\)-\(\frac{1}{10}\)+\(\frac{1}{10}\)-\(\frac{1}{17}\)+.....+\(\frac{1}{73}\)-\(\frac{1}{80}\)-\(\frac{1}{2}\)-\(\frac{1}{9}\)-......-\(\frac{1}{23}\)-\(\frac{1}{30}\))

P=\(\frac{1}{7}\)\(\times\)(\(\frac{1}{3}\)-\(\frac{1}{80}\))-\(\frac{1}{7}\)(\(\frac{1}{2}\)-\(\frac{1}{30}\))

P=\(\frac{1}{7}\)\(\times\)(\(\frac{1}{3}\)-\(\frac{1}{80}\)-\(\frac{1}{2}\)+\(\frac{1}{30}\))

P=\(\frac{-7}{336}\)

Bài này mk ko tính máy tính nên ko chắc đâu

26 tháng 2 2017

taị mk ko tính máy tính lên sai.

bn thông cảm nha. thường ngày hay dùng máy tính quá nên tính sai thì bn thông cảm

10 tháng 6 2017

a,        \(\frac{2^{30}.5^7+2^{13}.5^{27}}{2^{27}.5^7+2^{10}.5^{27}}=\frac{2^{13}.\left(2^{17}.5^7+5^{27}\right)}{2^{10}.\left(2^{17}.5^7+5^{27}\right)}=\frac{2^{13}}{2^{10}}=2^3=8\).

b,        \(\frac{81.2^2+3^4+20.9^2}{16.3^2+45+2^2.9}=\frac{3^4.2^2+3^4+20.3^4}{16.3^2+3^2.5+2^2.3^2}=\frac{3^4.\left(2^2+1+20\right)}{3^2.\left(16+5+2^2\right)}=\frac{3^4.25}{3^2.25}=\frac{3^4}{3^2}=3^2=9\)

10 tháng 6 2017

a : 8

b : 9

2 tháng 7 2017

đặt \(A=\frac{7}{10}+\frac{7}{10^2}+\frac{7}{10^3}+\frac{7}{10^4}\)

\(A=7.\left(\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+\frac{1}{10^4}\right)\)

Lại đặt \(B=\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+\frac{1}{10^4}\)

\(10B=1+\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}\)

\(10B-B=\left(1+\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}\right)-\left(\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+\frac{1}{10^4}\right)\)

\(9B=1-\frac{1}{10^4}\)

\(\Rightarrow B=\frac{1-\frac{1}{10^4}}{9}\)

\(\Rightarrow A=7.\frac{1-\frac{1}{10^4}}{9}=\frac{7.\left(1-\frac{1}{10^4}\right)}{9}\)

Nhưng có vô hạn số hạng thì sao bạn

21 tháng 3 2019

\(B=\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{\left(7\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\left(\frac{2}{7}\right)^2-\frac{4}{343}}\)

\(B=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}\)

\(B=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}\)

\(B=\frac{1}{4}\)

\(=\dfrac{1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}}{\dfrac{8}{2}-\dfrac{4}{7}+\dfrac{4}{49}-\dfrac{4}{343}}=\dfrac{1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}}{4-\dfrac{4}{7}+\dfrac{4}{49}-\dfrac{4}{343}}=\dfrac{1}{4}\)

HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023

a)

\(\begin{array}{l}A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right).\\A = \left( {\frac{{30}}{{15}} + \frac{5}{{15}} - \frac{6}{{15}}} \right) - \left( {\frac{{105}}{{15}} - \frac{9}{{15}} - \frac{{20}}{{15}}} \right) - \left( {\frac{3}{{15}} + \frac{{25}}{{15}} - \frac{{60}}{{15}}} \right)\\A = \frac{{29}}{{15}} - \frac{{76}}{{15}} - \left( {\frac{{ - 32}}{{15}}} \right)\\A = \frac{{29}}{{15}} - \frac{{76}}{{15}} + \frac{{32}}{{15}}\\A = \frac{{ - 15}}{{15}}\\A =  - 1\end{array}\)

b)

\(\begin{array}{l}A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right)\\A = 2 + \frac{1}{3} - \frac{2}{5} - 7 + \frac{3}{5} + \frac{4}{3} - \frac{1}{5} - \frac{5}{3} + 4\\A = \left( {2 - 7 + 4} \right) + \left( {\frac{1}{3} + \frac{4}{3} - \frac{5}{3}} \right) + \left( { - \frac{2}{5} + \frac{3}{5} - \frac{1}{5}} \right)\\A =  - 1 + 0 + 0 =  - 1\end{array}\)