Tính giá trị của biểu thức
20/2x7 + 20/7x12 + 20/12x17 + 20/17x22 + 20/22x27 + 20/27x32
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
26 tháng 6 2023
A = \(\dfrac{5}{1.6}\)+\(\dfrac{5}{6.11}\)+\(\dfrac{5}{11.16}\)+\(\dfrac{5}{16.21}\)+...+\(\dfrac{5}{101.106}\)
A = \(\dfrac{1}{1}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{101}-\dfrac{1}{106}\)
A = \(\dfrac{1}{1}\) - \(\dfrac{1}{106}\)
A = \(\dfrac{105}{106}\)
B = \(\dfrac{3}{1.4}\) +\(\dfrac{3}{4.7}\)+\(\dfrac{3}{7.10}\)+...+\(\dfrac{3}{97.100}\)
B = \(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\)
B = \(\dfrac{1}{1}\) - \(\dfrac{1}{100}\)
B = \(\dfrac{99}{100}\)
C = \(\dfrac{1}{2.7}+\dfrac{1}{7.12}\) + \(\dfrac{1}{12.17}\)+...+ \(\dfrac{1}{97.102}\)
C= \(\dfrac{1}{5}\) \(\times\)( \(\dfrac{5}{2.7}+\dfrac{5}{7.12}+\dfrac{5}{12.17}+...+\dfrac{5}{97.102}\))
C = \(\dfrac{1}{5}\)\(\times\)(\(\dfrac{1}{2}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{12}\) + \(\dfrac{1}{12}\) - \(\dfrac{1}{17}\)+...+ \(\dfrac{1}{97}\) - \(\dfrac{1}{102}\))
C = \(\dfrac{1}{5}\) \(\times\)( \(\dfrac{1}{2}\) - \(\dfrac{1}{102}\))
C = \(\dfrac{1}{5}\) \(\times\) \(\dfrac{25}{51}\)
C = \(\dfrac{5}{51}\)
D = \(\dfrac{1}{2}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\) + \(\dfrac{1}{30}\) + \(\dfrac{1}{42}\) + \(\dfrac{1}{56}\) + \(\dfrac{1}{72}\)
D = \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\)+\(\dfrac{1}{7.8}\)+ \(\dfrac{1}{8.9}\)
D = \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\)+\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\) - \(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{8}\)+\(\dfrac{1}{8}\)-\(\dfrac{1}{9}\)
D = \(\dfrac{1}{1}\) - \(\dfrac{1}{9}\)
D = \(\dfrac{8}{9}\)
E = \(\dfrac{3}{2.4}\)+\(\dfrac{3}{4.6}\)+\(\dfrac{3}{6.8}\)+...+\(\dfrac{3}{98.100}\)
E = \(\dfrac{3}{2}\) \(\times\) ( \(\dfrac{2}{2.4}\) + \(\dfrac{2}{4.6}\)+ \(\dfrac{2}{6.8}\)+...+\(\dfrac{2}{98.100}\))
E = \(\dfrac{3}{2}\)\(\times\)( \(\dfrac{1}{2}\) - \(\dfrac{1}{4}\)+ \(\dfrac{1}{4}\) - \(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{8}\)+...+\(\dfrac{1}{98}\) - \(\dfrac{1}{100}\))
E = \(\dfrac{3}{2}\) \(\times\) ( \(\dfrac{1}{2}\) - \(\dfrac{1}{100}\))
E = \(\dfrac{3}{2}\) \(\times\) \(\dfrac{49}{100}\)
E = \(\dfrac{147}{200}\)
DK
Tính giá trị của biểu thức sau : M = \(\frac{20}{112}+\frac{20}{280}+\frac{20}{542}+\frac{20}{832}\)
0
PT
0
HQ
Hà Quang Minh
Giáo viên
24 tháng 9 2023
Ta có: \(\sin {70^o} = \cos {20^o};\;\cos {110^o} = - \cos {70^o} = - \sin {20^o}\)
\(\begin{array}{l} \Rightarrow A = {(\sin {20^o} + \cos {20^o})^2} + {(\cos {20^o} - \sin {20^o})^2}\\ = ({\sin ^2}{20^o} + {\cos ^2}{20^o} + 2\sin {20^o}\cos {20^o}) + ({\cos ^2}{20^o} + {\sin ^2}{20^o} - 2\sin {20^o}\cos {20^o})\\ = 2({\sin ^2}{20^o} + {\cos ^2}{20^o})\\ = 2\end{array}\)
Ta có: \(\tan {110^o} = - \tan {70^o} = - \cot {20^o};\;\cot {110^o} = - \cot {70^o} = - \tan {20^o}.\)
\( \Rightarrow B = \tan {20^o} + \cot {20^o} + ( - \cot {20^o}) + ( - \tan {20^o}) = 0\)
Giúp với
Ta có:\(\dfrac{20}{2\times7}+\dfrac{20}{7\times12}+\dfrac{20}{12\times17}+\dfrac{20}{17\times22}+\dfrac{20}{22\times27}+\dfrac{20}{27\times32}\)
\(=4\times\left(\dfrac{5}{2\times7}+\dfrac{5}{7\times12}+\dfrac{5}{12\times17}+\dfrac{5}{17\times22}+\dfrac{5}{22\times27}+\dfrac{5}{27\times32}\right)\)
\(=4\times\left(\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{22}+\dfrac{1}{22}-\dfrac{1}{27}+\dfrac{1}{27}-\dfrac{1}{32}\right)\)
\(=4\times\left(\dfrac{1}{2}-\dfrac{1}{32}\right)=4\times\dfrac{15}{32}=\dfrac{30}{16}\)