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\(C=\frac{7^2}{2.9}+\frac{7^2}{9.16}+\frac{7^2}{16.23}+...+\frac{7^2}{65.72}\)
\(C=\frac{7^2}{7}.\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+\frac{1}{16}-\frac{1}{23}+...+\frac{1}{65}-\frac{1}{72}\right)\)
\(C=7.\left(\frac{1}{2}-\frac{1}{72}\right)\)
\(C=7.\frac{35}{72}=\frac{245}{72}\)
Ta có : \(C=\frac{7^2}{2.9}+\frac{7^2}{9.16}+\frac{7^2}{16.23}+.....+\frac{7^2}{65.72}\)
\(\Rightarrow C=7\left(\frac{7}{2.9}+\frac{7}{9.16}+\frac{7}{16.23}+.....+\frac{7}{65.72}\right)\)
\(\Rightarrow C=7\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+.....+\frac{1}{65}-\frac{1}{72}\right)\)
\(\Rightarrow C=7\left(\frac{1}{2}-\frac{1}{72}\right)\)
\(\Rightarrow C=7.\frac{35}{72}=\frac{245}{72}\)
\(\dfrac{7^2}{2.9}+\dfrac{7^2}{9.16}+\dfrac{7^2}{16.23}+...+\dfrac{7^2}{65.72}\)
\(=7^2\left(\dfrac{1}{2.9}+\dfrac{1}{9.16}+\dfrac{1}{16.23}+...+\dfrac{1}{65.72}\right)\)
\(=7^2\left(\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{23}+...+\dfrac{1}{65}-\dfrac{1}{72}\right)\)
\(=7^2\left(\dfrac{1}{2}-\dfrac{1}{72}\right)\)
\(=49\left(\dfrac{35}{72}\right)\)
\(=\dfrac{1715}{72}\)
\(l=\dfrac{7^2}{2.9}+\dfrac{7^2}{9.16}+\dfrac{7^2}{16.23}+...+\dfrac{7^2}{65.72}\)
\(=7\left(\dfrac{7}{2.9}+\dfrac{7}{9.16}+\dfrac{7}{16.23}+...+\dfrac{7}{65.72}\right)\)
\(=7\left(\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{23}+...+\dfrac{1}{65}-\dfrac{1}{72}\right)\)
\(=7\left(\dfrac{1}{2}-\dfrac{1}{72}\right)=7\left(\dfrac{36}{72}-\dfrac{1}{72}\right)=7.\dfrac{35}{72}=\dfrac{245}{72}\)
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
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
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, \(\frac{81.2^2+3^4+20.9^2}{16.3^2+45+2^2.9}=\frac{3^4.2^2+3^4+20.3^4}{16.3^2+3^2.5+2^2.3^2}=\frac{3^4.\left(2^2+1+20\right)}{3^2.\left(16+5+2^2\right)}=\frac{3^4.25}{3^2.25}=\frac{3^4}{3^2}=3^2=9\)
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 \(B=\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+\frac{1}{10^4}\)
\(10B=1+\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}\)
\(10B-B=\left(1+\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}\right)-\left(\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+\frac{1}{10^4}\right)\)
\(9B=1-\frac{1}{10^4}\)
\(\Rightarrow B=\frac{1-\frac{1}{10^4}}{9}\)
\(\Rightarrow A=7.\frac{1-\frac{1}{10^4}}{9}=\frac{7.\left(1-\frac{1}{10^4}\right)}{9}\)
Nếu phân số thứ 2 là \(\frac{1}{10.17}\) thì làm như vậy nè
\(\frac{1}{3.10}+\frac{1}{10.17}+...+\frac{1}{73.80}-\frac{1}{2.9}-\frac{1}{9.16}-\frac{1}{16.23}-\frac{1}{23.30}\)
= \(\frac{1}{7}\left(\frac{1}{3}-\frac{1}{10}+\frac{1}{10}-\frac{1}{17}+...+\frac{1}{73}-\frac{1}{80}\right)-\left(\frac{1}{2.9}+\frac{1}{9.16}+\frac{1}{16.23}+\frac{1}{23.30}\right)\)
= \(\frac{1}{7}\left(\frac{1}{3}-\frac{1}{80}\right)-\frac{1}{7}\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+\frac{1}{16}-\frac{1}{23}+\frac{1}{23}-\frac{1}{30}\right)\)
= \(\frac{1}{7}.\frac{77}{240}-\frac{1}{7}\left(\frac{1}{2}-\frac{1}{30}\right)=\frac{1}{7}.\frac{77}{240}-\frac{1}{7}.\frac{7}{15}\)
= \(\frac{11}{240}-\frac{1}{15}\)
= \(-\frac{1}{48}\)
\(B=\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{\left(7\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\left(\frac{2}{7}\right)^2-\frac{4}{343}}\)
\(B=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}\)
\(B=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}\)
\(B=\frac{1}{4}\)
\(=\dfrac{1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}}{\dfrac{8}{2}-\dfrac{4}{7}+\dfrac{4}{49}-\dfrac{4}{343}}=\dfrac{1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}}{4-\dfrac{4}{7}+\dfrac{4}{49}-\dfrac{4}{343}}=\dfrac{1}{4}\)
\(C=\frac{7^2}{2.9}+\frac{7^2}{9.16}+\frac{7^2}{16.23}+....+\frac{7^2}{65.72}\)
\(C=\frac{7^2}{7}\cdot\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+\frac{1}{16}-\frac{1}{23}+....+\frac{1}{65}-\frac{1}{72}\right)\)
\(C=7\cdot\left(\frac{1}{2}-\frac{1}{72}\right)\)
\(C=7\cdot\frac{35}{72}=\frac{245}{72}\)
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