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Ta có
\(2017-\left(\frac{1}{4}+\frac{2}{5}+\frac{3}{6}+\frac{4}{7}+...+\frac{2017}{2020}\right)\)
\(=\left(1+1+...+1\right)-\left(\frac{1}{4}+\frac{2}{5}+...+\frac{2017}{2020}\right)\)
\(=\left(1-\frac{1}{4}\right)+\left(1-\frac{2}{5}\right)+...+\left(1-\frac{2017}{2020}\right)\)
\(=\frac{3}{4}+\frac{3}{5}+....+\frac{3}{2020}\)
\(=\frac{3.5}{4.5}+\frac{3.5}{5.5}+\frac{3.5}{6.5}+...+\frac{3.5}{2020.5}\)
\(=3.5\left(\frac{1}{4.5}+\frac{1}{5.5}+\frac{1}{6.5}+...+\frac{1}{2020.5}\right)\)
\(=15.\left(\frac{1}{20}+\frac{1}{25}+\frac{1}{30}+...+\frac{1}{10100}\right)\)
Thế vào ta có
\(\frac{15.\left(\frac{1}{20}+\frac{1}{25}+\frac{1}{30}+...+\frac{1}{10100}\right)}{\frac{1}{20}+\frac{1}{25}+...+\frac{1}{10100}}=15\)
Được cập nhật 41 giây trước (17:23)
Ta có :
2017−(14 +25 +36 +47 +...+20172020 )
=(1+1+...+1)−(14 +25 +...+20172020 )
=(1−14 )+(1−25 )+...+(1−20172020 )
=34 +35 +....+32020
=3.54.5 +3.55.5 +3.56.5 +...+3.52020.5
=3.5(14.5 +15.5 +16.5 +...+12020.5 )
=15.(1
C\(\frac{1}{1}-\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+\frac{1}{5.6}\)-\(\frac{1}{6.7}\)+\(\frac{1}{7.8}\)-\(\frac{1}{8.9}+\frac{1}{9.10}\)
c=\(\frac{1}{1}-\frac{1}{10}\)
c=\(\frac{9}{10}\)
còn a và b rễ lắm mình ko thích làm bài rễ đâu bạn cố chờ lời giải khác nhé!
\(A=\frac{15\times3^{11}+4\times27^4}{9^7}\)
\(A=\frac{15\times177147+4\times531441}{4782969}\)
\(A=\frac{2657205+2125764}{4782969}\)
\(A=\frac{47829969}{47829969}=1\)
Tờ làm luôn, ko ghi đề nữa nhé
\(A=\frac{\frac{24}{12}-\frac{4}{12}+\frac{3}{12}}{\frac{24}{12}+\frac{2}{12}-\frac{3}{12}}\)
\(A=\frac{\frac{23}{12}}{\frac{23}{12}}=1\)
Vậy A=1
\(A=\frac{2-\frac{1}{3}+\frac{1}{4}}{2+\frac{1}{6}-\frac{1}{4}}\)\(=\frac{2-\frac{2}{6}+\frac{2}{8}}{2+\frac{2}{12}-\frac{2}{8}}\)\(=\frac{2\left(1-\frac{1}{6}+\frac{1}{8}\right)}{-2\left(1-\frac{1}{12}+\frac{1}{8}\right)}\)\(=-1\)
\(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right].x=\frac{9}{1}+\frac{8}{2}+...+\frac{1}{9}\)
=> \(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right].x=\frac{10-1}{1}+\frac{10-2}{2}+...+\frac{10-9}{9}\)
=> \(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right].x=\frac{10}{1}-1+...+\frac{10}{9}-1\)
=> \(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right]x=10-9+\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}\)= \(\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}+\frac{10}{10}\)
=>\(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right]x=10\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)\)
=> \(x=10\)
b) Tương tự câu a
a) \(A=\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+......+\frac{1}{2017.2022}\)
\(5A=5.\left(\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+.....+\frac{1}{2017.2022}\right)\)
\(5A=\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+......+\frac{5}{2017.2022}\)
\(5A=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+........+\frac{1}{2017}-\frac{1}{2022}\)
\(5A=1-\frac{1}{2022}\)
\(5A=\frac{2022}{2022}-\frac{1}{2022}\)
\(5A=\frac{2021}{2022}\)
\(A=\frac{2021}{2022}\div5\)
\(A=\frac{20201}{10110}\)
TL:
\(\frac{5}{6}=\frac{1}{2}+\frac{1}{3}\)
@@@@@@@@@@
HT
\(M=1+\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+...+\frac{1}{3^{19}}-\frac{1}{3^{20}}\)
đặt \(A=\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+...+\frac{1}{3^{19}}-\frac{1}{3^{20}}\)
\(3A=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{18}}-\frac{1}{3^{19}}\)
\(4A=1-\frac{1}{3^{20}}\)
\(A=\frac{1-\frac{1}{3^{20}}}{4}\)
\(M=1+\frac{1-\frac{1}{3^{20}}}{4}=\frac{5-\frac{1}{3^{20}}}{4}\)
Ta có : 1:M=1+3-3^2+3^3-3^4+....+3^19-3^20
1/M=(1+3^2+3^4+....3^20)-(3+3^3+..+3^19)
1/M=[(3^20-1)/8]-[(3^21-3)/8]
1/M=[3^20-3^21+(-2)]/8
Bạn tự làm tiếp nhé
1/1015
chắc chắn rồi bn ạ
mk k bik làm
ai làm hộ mk đi