c=12^2+14^2+16^2+......+40^2
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LH
Lê Huy Tường
VIP
15 tháng 6 2021
2-4+6-8+10-12+14-16+18-20+22
=(2+18)-(4+16)+(6+14)-(8+12)-(10+20)+22
=20-20+20-20-20+22
=22-20
=2
NH
2
21 tháng 6 2018
\(E=\frac{2}{10x12}+\frac{2}{12x14}+...+\frac{2}{108x110}\)
\(\Rightarrow E=\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+...+\frac{1}{108}-\frac{1}{110}\)
\(\Rightarrow E=\frac{1}{10}-\frac{1}{110}=\frac{10}{110}=\frac{1}{11}\)
Công thức :\(\frac{a}{n\left(n+a\right)}=\frac{1}{n}-\frac{1}{n+a}\)
21 tháng 6 2018
\(E=\frac{2}{10\cdot12}+\frac{2}{12\cdot14}+\frac{2}{14\cdot16}+...+\frac{2}{108\cdot110}\)
\(E=\frac{1}{10\cdot12}+\frac{1}{12\cdot14}+\frac{1}{14\cdot16}+...+\frac{1}{108\cdot110}\)
\(E=\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+\frac{1}{14}-\frac{1}{16}+...+\frac{1}{108}-\frac{1}{110}\)
\(E=\frac{1}{10}-\frac{1}{110}\)
\(E=\frac{11}{110}-\frac{1}{110}\)
\(E=\frac{10}{110}\)\(=\frac{1}{11}\)
Học tốt nha bạn !!!
16 tháng 5 2019
\(a.\frac{27.45+27.55}{2+4+6+...+14+16+18}=\frac{27.100}{\frac{\left(2+18\right).9}{2}}=30\)
\(b.\frac{26.108-26.12}{32-28+24-20+16-12+8-4}=\frac{26\left(108-12\right)}{\left(32-28\right).4}=\frac{26.96}{4.4}=156\)
\(c.\frac{27.4500+135.550.2}{2+4+6+...+14+16+18}=\frac{270.450+270.550}{\frac{\left(2+8\right).9}{2}}\)
\(d.\frac{48.700-24.45.20}{45-40+35-30+25-20+15-10+5}=\frac{48.700-48.450}{5.5}\)\(=\frac{48\left(700-450\right)}{25}=\frac{48.250}{25}=480\)
#ĐinhBa
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
27 tháng 7
a; \(\dfrac{1}{4}\) + \(\dfrac{2}{5}\) + \(\dfrac{6}{8}\) + \(\dfrac{9}{15}\) + \(\dfrac{8}{1}\)
= (\(\dfrac{1}{4}\) + \(\dfrac{6}{8}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{9}{15}\)) + \(\dfrac{8}{1}\)
= (\(\dfrac{1}{4}\) + \(\dfrac{3}{4}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{3}{5}\)) + 8
= 1 + 1 + 8
= 2 + 8
= 10
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
27 tháng 7
b; \(\dfrac{1}{2}\) + \(\dfrac{2}{4}\) + \(\dfrac{3}{6}\) + \(\dfrac{4}{8}\) + \(\dfrac{5}{10}\) + \(\dfrac{6}{12}\) + \(\dfrac{7}{14}\) + \(\dfrac{8}{16}\) + \(\dfrac{10}{20}\)
= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) x (\(\dfrac{2}{2}\) + \(\dfrac{3}{3}\) + \(\dfrac{4}{4}\) + \(\dfrac{5}{5}\)+ \(\dfrac{6}{6}+\dfrac{7}{7}+\dfrac{8}{8}\) + \(\dfrac{10}{10}\))
= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) x (1 + 1 +1 + 1+ 1+ 1+ 1 +1)
= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) x 1 x 8
= \(\dfrac{1}{2}\) + \(\)\(\dfrac{1}{2}\) x 8
= \(\dfrac{1}{2}\) + 4
= \(\dfrac{9}{2}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
21 tháng 9
a; \(\dfrac{1}{4}\) + \(\dfrac{2}{5}\) + \(\dfrac{6}{8}\) + \(\dfrac{9}{15}\) + \(\dfrac{8}{1}\)
= (\(\dfrac{1}{4}\) + \(\dfrac{6}{8}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{9}{15}\)) + 8
= (\(\dfrac{1}{4}\) + \(\dfrac{3}{4}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{3}{5}\)) + 8
= 1 + 1 + 8
= 2 + 8
= 10
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
21 tháng 9
b; \(\dfrac{1}{2}\) + \(\dfrac{2}{4}\) + \(\dfrac{3}{6}\) + \(\dfrac{4}{8}\) + \(\dfrac{5}{10}\) + \(\dfrac{6}{12}\) + \(\dfrac{7}{14}\) + \(\dfrac{8}{16}\) + \(\dfrac{9}{18}\) + \(\dfrac{10}{20}\)
= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\)
= \(\dfrac{1}{2}\) x 10
= 5
XV
0
TV
2
H9
HT.Phong (9A5)
CTVHS
21 tháng 3 2023
\(A=\dfrac{2}{20.12}+\dfrac{2}{12.14}+\dfrac{2}{14.16}+...+\dfrac{2}{98.100}\)
\(A=\dfrac{1}{10}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{16}+...+\dfrac{1}{98}-\dfrac{1}{100}\)
\(A=\dfrac{1}{10}-\dfrac{1}{100}\)
\(A=\dfrac{9}{100}\)
BC
1
10 tháng 7 2017
\(A⋮8\)( vì 2*4*6*8*10*12*14*16\(⋮8\)và \(40⋮8\))
A ko chia hết 6 vì (2*4*6*8*10*12*14*16\(⋮6\)mà 40 ko chia hết )
tương tự với 5
c = 12² + 14² + 16² + ... + 40²
= (2.6)² + (2.7)² + (2.8)² + ... + (2.20)²
= 2².6² + 2².7² + 2².8² + ... + 2².20²
= 4.(6² + 7² + 8² + ... + 20²)
= 4.(1² + 2² + 3² + 4² + ... + 20²) - 4.(1² + 2² + 3² + 4² + 5²)
= 4.20.(20 + 1).(2.20 + 1) : 6 - 4.5.(5 + 1).(2.5 + 1) : 6
= 11480 - 220
= 11260
mjjnn n