Tính bằng cách hợp lí
S= 1/5+1/20+1/44+1/77+1/119
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NT
3
5 tháng 7 2015
Bài 1 :
Đặt A=1.2+2.3+3.4+4.5+.........+99.100
=> 3A=1.2.3+2.3.(4-1)+........+99.100.(101-98)
3A=1.2.3+2.3.4-1.2.3+........+99.100.101-98.99.100
3A=99.100.101
A=33.100.101
A=333300
Bài 2 :
1:20 + 1:44 + 1:77 + 1:119 + 1:170 = \(\frac{1}{20}+\frac{1}{44}+\frac{1}{77}+\frac{1}{119}+\frac{1}{170}=\frac{1}{10}=0,1\)
5 tháng 7 2015
1)1.2+2.3+3.4+4.5+...+99.100
đặt 3D=1.2+2.3+3.4+...+99.100
=1.2.3+2.8.3+...+3.4.3+4.5.3+...+99.100.3
=1.2.3+2.3.(4-1)+3.4.(5-2)+4.5.(6-3)+...+99.100.(101-98)
=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5
=99.100.101
=999900
D=999900:3=333300
nếu đúng nhớ cảm ơn nhak. mình ko bít làm bài 2
DD
Đoàn Đức Hà
Giáo viên
14 tháng 7 2021
\(A=\frac{2}{5.7}+\frac{5}{7.12}+\frac{7}{12.19}+\frac{9}{19.28}+\frac{11}{28.39}+\frac{1}{39.40}\)
\(=\frac{7-5}{5.7}+\frac{12-7}{7.12}+\frac{19-12}{12.19}+\frac{28-19}{19.28}+\frac{39-28}{28.39}+\frac{40-39}{39.40}\)
\(=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{19}+\frac{1}{19}-\frac{1}{28}+\frac{1}{28}-\frac{1}{39}+\frac{1}{39}-\frac{1}{40}\)
\(=\frac{1}{5}-\frac{1}{40}=\frac{7}{40}\)
\(B=\frac{1}{20}+\frac{1}{44}+\frac{1}{77}+\frac{1}{119}+\frac{1}{170}\)
\(=\frac{2}{40}+\frac{2}{88}+\frac{2}{154}+\frac{2}{238}+\frac{2}{340}\)
\(=\frac{2}{5.8}+\frac{2}{8.11}+\frac{2}{11.14}+\frac{2}{14.17}+\frac{2}{17.20}\)
\(=\frac{2}{3}\left(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\right)\)
\(=\frac{2}{3}\left(\frac{8-5}{5.8}+\frac{11-8}{8.11}+\frac{14-11}{11.14}+\frac{17-14}{14.17}+\frac{20-17}{17.20}\right)\)
\(=\frac{2}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\right)\)
\(=\frac{2}{3}\left(\frac{1}{5}-\frac{1}{20}\right)=\frac{1}{10}\)
\(\frac{A}{B}=\frac{\frac{7}{40}}{\frac{1}{10}}=\frac{7}{4}\)
NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 2 2021
Câu C giải rồi
\(B=\dfrac{1}{5}+\dfrac{1}{20}+\dfrac{1}{44}+\dfrac{1}{77}+\dfrac{1}{119}+\dfrac{1}{170}+\dfrac{1}{230}+\dfrac{1}{299}\)
\(=2\left(\dfrac{1}{10}+\dfrac{1}{40}+\dfrac{1}{88}+\dfrac{1}{154}+\dfrac{1}{238}+\dfrac{1}{340}+\dfrac{1}{460}+\dfrac{1}{598}\right)\)
\(=\dfrac{2}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+\dfrac{3}{14.17}+\dfrac{3}{17.20}+\dfrac{3}{20.23}+\dfrac{3}{23.26}\right)\)
\(=\dfrac{2}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{23}-\dfrac{1}{26}\right)\)
\(=\dfrac{2}{3}\left(\dfrac{1}{2}-\dfrac{1}{26}\right)=\dfrac{4}{13}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
20 tháng 8 2023
A = \(\dfrac{2}{5.7}\) + \(\dfrac{5}{7.12}\) + \(\dfrac{7}{12.19}\) + \(\dfrac{9}{19.28}\) + \(\dfrac{11}{28.39}\) + \(\dfrac{1}{30.40}\)
A = \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{12}\) + \(\dfrac{1}{12}\) - \(\dfrac{1}{19}\) + \(\dfrac{1}{19}\) - \(\dfrac{1}{28}\) + \(\dfrac{1}{28}\) - \(\dfrac{1}{39}\) + \(\dfrac{1}{1200}\)
A = \(\dfrac{1}{5}\) - \(\dfrac{1}{39}\) + \(\dfrac{1}{1200}\)
A = \(\dfrac{34}{195}\) + \(\dfrac{1}{1200}\)
B = \(\dfrac{1}{20}\) + \(\dfrac{1}{44}\) + \(\dfrac{1}{77}\) + \(\dfrac{1}{119}\) + \(\dfrac{1}{170}\)
B = 2 \(\times\) ( \(\dfrac{1}{2.20}\) + \(\dfrac{1}{2.44}\) + \(\dfrac{1}{2.77}\) + \(\dfrac{1}{2.119}\) + \(\dfrac{1}{2.170}\))
B = 2 \(\times\) ( \(\dfrac{1}{40}\) + \(\dfrac{1}{88}\) + \(\dfrac{1}{154}\) + \(\dfrac{1}{238}\) + \(\dfrac{1}{340}\))
B = 2 \(\times\) ( \(\dfrac{1}{5.8}\) + \(\dfrac{1}{8.11}\) + \(\dfrac{1}{11.14}\) + \(\dfrac{1}{14.17}\) + \(\dfrac{1}{17.20}\))
B = \(\dfrac{2}{3}\) \(\times\) ( \(\dfrac{3}{5.8}\) + \(\dfrac{3}{8.11}\)+ \(\dfrac{3}{11.14}\) + \(\dfrac{3}{14.17}\) + \(\dfrac{3}{17.20}\))
B = \(\dfrac{2}{3}\) \(\times\) ( \(\dfrac{1}{5}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\) - \(\dfrac{1}{11}\) + \(\dfrac{1}{11}\) - \(\dfrac{1}{14}\) + \(\dfrac{1}{14}\) - \(\dfrac{1}{17}\) + \(\dfrac{1}{17}\) - \(\dfrac{1}{20}\))
B = \(\dfrac{2}{3}\) \(\times\) ( \(\dfrac{1}{5}\) - \(\dfrac{1}{20}\))
B = \(\dfrac{2}{3}\) \(\times\) \(\dfrac{3}{20}\)
B = \(\dfrac{1}{10}\) = \(\dfrac{34}{340}\) < \(\dfrac{34}{195}\) + \(\dfrac{1}{1200}\)
Vậy A > B
NV
30 tháng 4 2023
Ta có : +) A= 1/5 -1/7 +1/7 -1/12 +1/12 - 1/19 +1/19 - 1/28 +1/28 - 1/39 +1/30.40 ⇔ A=1/5 -1/39 +1/30.40
+) B= 2.(1/5.8 +1/8.11 +1/11.14 +1/14.17 + 1/17.20 )
⇔B=2. 1/3.(1/5 - 1/8 +1/8 - 1/11 +1/11- 1/14 +1/14 -1/17 +1/17 -1/20 )
⇔B=2/3.( 1/5-1/20 ) Ta luôn có :B luôn <1/5 - 1/20
Mà 1/5 -1/20 <1/5 -1/39 +1/30.40 =A
⇒ A>B (dpcm) Tích mình với nha bn .
12 tháng 5 2023
Sửa đề: 39*40
\(A=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{12}+...+\dfrac{1}{39}-\dfrac{1}{40}=\dfrac{1}{5}-\dfrac{1}{40}=\dfrac{7}{40}\)
\(B=\dfrac{2}{3}\left(\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{17\cdot20}\right)\)
\(=\dfrac{2}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{17}-\dfrac{1}{20}\right)\)
=2/3*3/20=2/20=1/10=4/40<A
ai làm nhanh đầy đủ mk k cho
Nó dài lắm bạn à!!!!
Bó tay luôn!!!!
Đúng thì k nha!!!!