Tinh 6/7x12 + 6/12x17 + ...... +6/87 x 92 + 6/92 x95
mình se ra cau do thuong xuyen
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em ơi, em đdặt câu hỏi sai rồi, Ko có số nào chia hết cho 10 đc đâu. Nhưng 10 có thể chia đc 7 số trên nhưng là số thập phân( số có dấu phẩy )( mai sau lớp 5 thì em biết)
56+87+59+92+84+93+35=506 chia 10 dư 6
=>Muốn tìm tổng6 số chia hết cho 10 thì phải lấy tổng của 7 số này trừ đi số mà chia cho 10 có cùng số dư với tổng của 7 số
=>Số cần loại đi là 56
=>Tổng sẽ là 506-56=450
a) 127 + (-18) + (-107) + (-92)
=(-18 + -92) + 127 +(-107)
=(-37) + 127 + (-107)
=90 + (-107)
= -17
b) x- 3) -7= -4
x- 3 = (-4) + 7
x- 3 = 3
x = 3 + 3
x = 6
Vậy x = 6
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 = \(\dfrac{147}{200}\)
\(42+\frac{93}{6}+\frac{87}{12}+\frac{79}{20}+\frac{69}{30}+\frac{57}{42}\)
\(=37+\left(1+\frac{93}{6}\right)+\left(1+\frac{87}{12}\right)+\left(\frac{79}{20}+1\right)+\left(1+\frac{69}{30}\right)+\left(1+\frac{57}{42}\right)\)
\(=37+\frac{99}{6}+\frac{99}{12}+\frac{99}{20}+\frac{99}{30}+\frac{99}{42}\)
\(=37+99\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)
\(=37+99\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(=37+99\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(=37+99\left(\frac{1}{2}-\frac{1}{7}\right)\)
\(=37+99.\frac{5}{14}\)
\(=37+35\frac{5}{14}\)
\(=72\frac{5}{14}\)
\(\dfrac{92}{6}\times\dfrac{72}{15}:\left(-\dfrac{72}{10}\right)\)
\(=\dfrac{92\times72}{6\times15}:\left(-\dfrac{72}{10}\right)\)
\(=\dfrac{368}{5}:\left(-\dfrac{72}{10}\right)\)
\(=\dfrac{368}{5}\times\dfrac{-10}{72}\)
\(=\dfrac{368\times\left(-10\right)}{5\times72}\)
\(=\dfrac{-92}{9}\)
\(=\dfrac{92}{6}\cdot\dfrac{72}{15}\cdot\dfrac{-10}{72}=\dfrac{92}{6}\cdot\dfrac{-2}{3}=\dfrac{-92}{9}\)
dua thoi gap gap lam mình can so sánh kẹt qua
\(\frac{6}{7.12}+\frac{6}{12.17}+...+\frac{6}{87.92}+\frac{6}{92.95}\)
= \(6\left(\frac{5}{7.12}.\frac{1}{5}+\frac{5}{12.17}.\frac{1}{5}+...+\frac{5}{92.95}.\frac{1}{5}\right)\)
= \(6.\frac{1}{5}\left(\frac{5}{7.12}+\frac{5}{12.17}+...+\frac{5}{87.92}+\frac{5}{92.95}\right)\)
= \(\frac{6}{5}\left(\frac{5}{7}-\frac{5}{12}+\frac{5}{12}-\frac{5}{17}+...+\frac{5}{92}-\frac{5}{95}\right)\)
= \(\frac{6}{5}\left(\frac{5}{7}-\frac{5}{95}\right)\)= \(\frac{6}{5}.\frac{88}{133}=\frac{528}{665}\)
Tự rút gọn, mình lười.