: 3/2x7 +3/7x12+...+3/97x102
tính x là nhân nha ko phải tìm X
K
Khách
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B
4
31 tháng 8 2023
\(=\dfrac{7-2}{2.7}+\dfrac{12-7}{7.12}+...+\dfrac{102-97}{97.102}\)
\(=\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{12}+...+\dfrac{1}{97}-\dfrac{1}{102}\)
\(=\dfrac{1}{2}-\dfrac{1}{102}=\dfrac{25}{51}\)
31 tháng 8 2023
\(=\dfrac{7-2}{2.7}+\dfrac{12-7}{7.12}+...+\dfrac{97-92}{92.97}\)
\(=\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{12}+...+\dfrac{1}{92}-\dfrac{1}{97}\)
\(=\dfrac{1}{2}-\dfrac{1}{97}=\dfrac{95}{194}\)
NH
Tính thuận giúp 4/3x7 + 5/7x12 + 1/12x13 + 7/13x20 + 3/20x 23
chú ý : x là chữ x không phải dấu nhân
0
NX
22 tháng 8 2023
a) \(215+x=400\)
\(\Rightarrow x=400-215\)
\(\Rightarrow x=185\)
b) \(12,5-2x=\dfrac{1}{5}\)
\(\Rightarrow2x=12,5-\dfrac{1}{5}\)
\(\Rightarrow2x=\dfrac{123}{10}\)
\(\Rightarrow x=\dfrac{123}{10}:2\)
\(\Rightarrow x=\dfrac{123}{20}\)
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 = \(\dfrac{147}{200}\)
KA
5
4 tháng 10 2016
\(4\times x^3+15=47\)
\(4\times x^3=32\)
\(x^3=8\)
\(x=2\)
DT
2
DP
1
27 tháng 2 2023
\(=5+\dfrac{4}{5}\left(\dfrac{5}{2\cdot7}+\dfrac{5}{7\cdot12}+...+\dfrac{5}{52\cdot57}\right)\)
\(=5+\dfrac{4}{5}\left(\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{12}+...+\dfrac{1}{52}-\dfrac{1}{57}\right)\)
\(=5+\dfrac{4}{5}\cdot\dfrac{55}{114}=\dfrac{307}{57}\)
Y
16 tháng 4 2022
\(a,\dfrac{7}{12}-\left(x+\dfrac{7}{10}\right):\dfrac{6}{5}=\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{7}{12}-x-\dfrac{7}{10}:\dfrac{6}{5}=\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{7}{12}-x-\dfrac{7}{12}=\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{7}{12}-x=\dfrac{5}{4}+\dfrac{7}{12}\)
\(\Leftrightarrow\dfrac{7}{12}-x=\dfrac{11}{6}\)
\(\Leftrightarrow x=\dfrac{7}{12}-\dfrac{11}{6}\)
\(\Leftrightarrow\dfrac{-5}{4}\)
\(\frac{3}{2.7}+\frac{3}{7.12}+\frac{3}{97.102}=\frac{3}{5}\left(\frac{5}{2.7}+\frac{5}{7.12}+...+\frac{9}{97.102}\right)\)
\(=\frac{3}{5}\left(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+...+\frac{1}{97}-\frac{1}{102}\right)=\frac{3}{5}\left(\frac{1}{2}-\frac{1}{102}\right)=\frac{3}{5}.\frac{25}{51}=\frac{5}{17}\)
5/17 nha bn