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VN
Võ Ngọc Phương
VIP
6 tháng 9 2023
\(\dfrac{1}{7}+\dfrac{1}{91}+\dfrac{1}{247}+\dfrac{1}{475}+\dfrac{1}{775}+\dfrac{1}{1147}\)
\(=\dfrac{1}{1.7}+\dfrac{1}{7.13}+\dfrac{1}{13.19}+\dfrac{1}{19.25}+\dfrac{1}{25.31}+\dfrac{1}{31.37}\)
\(=\dfrac{1}{6}\left(1-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{31}+\dfrac{1}{31}-\dfrac{1}{37}\right)\)
\(=\dfrac{1}{6}\left(1-\dfrac{1}{37}\right)\)
\(=\dfrac{1}{6}.\dfrac{36}{37}\)
\(=\dfrac{6}{37}\)
\(#Wendy.Dang\)
NT
1
H
15 tháng 4 2019
\(\frac{-1}{91}+\frac{-1}{247}+\frac{-1}{475}+\frac{-1}{775}+\frac{-1}{1147}\)
\(=-\left(\frac{1}{7.13}+\frac{1}{13.19}+\frac{1}{19.25}+\frac{1}{25.31}+\frac{1}{31.37}\right)\)
\(=-[\frac{1}{6}.\left(\frac{6}{7.13}+\frac{6}{13.19}+\frac{6}{19.25}+\frac{6}{25.31}+\frac{6}{31.37}\right)]\)
\(=-\text{[}\frac{1}{6}.\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{19}+\frac{1}{19}-\frac{1}{25}+\frac{1}{25}-\frac{1}{31}+\frac{1}{31}-\frac{1}{37}\right)\text{]}\)
\(=-\text{[}\frac{1}{6}.\left(\frac{1}{7}-\frac{1}{37}\right)\text{]}\)
\(=-\text{[}\frac{1}{6}.\frac{30}{259}\text{]}\)
\(=-\frac{5}{259}\)
T
26 tháng 9 2023
\(\dfrac{1}{7}+\dfrac{1}{91}+...+\dfrac{1}{1147}\)
\(=\dfrac{1}{1\cdot7}+\dfrac{1}{7\cdot13}+\dfrac{1}{13\cdot19}+...+\dfrac{1}{31\cdot37}\)
\(=\dfrac{1}{6}\cdot\left(\dfrac{6}{1\cdot7}+\dfrac{6}{7\cdot13}+\dfrac{6}{13\cdot19}+...+\dfrac{6}{31\cdot37}\right)\)
\(=\dfrac{1}{6}\cdot\left(1-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{19}+...+\dfrac{1}{31}-\dfrac{1}{37}\right)\)
\(=\dfrac{1}{6}\cdot\left(1-\dfrac{1}{37}\right)\)
\(=\dfrac{1}{6}\cdot\dfrac{36}{37}\)
\(=\dfrac{6}{37}\)
Vậy ...
#\(Toru\)
PL
2
H9
HT.Phong (9A5)
CTVHS
9 tháng 11 2023
\(\left(x-1\right)^3-\left(\dfrac{2}{2023}-\dfrac{7}{247}+\dfrac{1}{8}\right)=\dfrac{7}{247}-\dfrac{2}{2023}\)
\(\Rightarrow\left(x-1\right)^3-\dfrac{2}{2023}+\dfrac{7}{247}-\dfrac{1}{8}=\dfrac{7}{247}-\dfrac{2}{2023}\)
\(\Rightarrow\left(x-1\right)^3=\dfrac{7}{247}-\dfrac{7}{247}-\dfrac{2}{2023}+\dfrac{2}{2023}+\dfrac{1}{8}\)
\(\Rightarrow\left(x-1\right)^3=\dfrac{1}{8}\)
\(\Rightarrow\left(x-1\right)^3=\left(\dfrac{1}{2}\right)^3\)
\(\Rightarrow x-1=\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{2}+1\)
\(\Rightarrow x=\dfrac{3}{2}\)
AH
Akai Haruma
Giáo viên
9 tháng 11 2023
Lời gải:
$(x-1)^3=\frac{7}{247}-\frac{2}{2023}+\frac{2}{2023}-\frac{7}{247}+\frac{1}{8}=\frac{1}{8}$
$x-1=\frac{1}{2}$
$x=\frac{1}{2}+1=\frac{3}{2}$
BD
0
VP
2
24 tháng 8 2015
thế này à:
\(\frac{91-\frac{1}{11}-\frac{2}{12}-\frac{3}{13}-...-\frac{91}{101}}{\frac{1}{55}+\frac{1}{60}+....+\frac{1}{505}}\)
HT
24 tháng 8 2015
\(\frac{91-\frac{1}{11}-\frac{2}{12}-\frac{3}{13}-...-\frac{91}{101}}{\frac{1}{55}+\frac{1}{60}+\frac{1}{65}+...+\frac{1}{505}}\)
Xét tử:
\(91-\frac{1}{11}-\frac{2}{12}-\frac{3}{13}-...-\frac{91}{101}\)
= \(\left(1+1+1+...+1\right)-\left(\frac{1}{11}+\frac{2}{12}+\frac{3}{13}+...+\frac{91}{101}\right)\)
= \(\left(1-\frac{1}{11}\right)+\left(1-\frac{2}{12}\right)+....+\left(1-\frac{91}{101}\right)\)
= \(\frac{10}{11}+\frac{10}{12}+...+\frac{10}{101}\)
= \(10.\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{101}\right)\)
= \(10.5.\left(\frac{1}{55}+\frac{1}{60}+...+\frac{1}{505}\right)\)
= \(50.\left(\frac{1}{55}+\frac{1}{60}+...+\frac{1}{505}\right)\)
Thay vào ta được phân số:
\(\frac{50.\left(\frac{1}{55}+\frac{1}{60}+...+\frac{1}{505}\right)}{\frac{1}{55}+\frac{1}{60}+...+\frac{1}{505}}\)
= 50
TQ
0