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C = \(\frac{3}{2.3.4}+\frac{3}{3.4.5}+.....+\frac{3}{98.99.100}\)
C = \(3.\left(\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\right)\)
C = \(3.\frac{1}{2}.\left(\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\right)\)
C = \(\frac{3}{2}.\left(\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{100-98}{98.99.100}\right)\)
C = \(\frac{3}{2}.\left(\frac{4}{2.3.4}-\frac{2}{2.3.4}+\frac{5}{3.4.5}-\frac{3}{3.4.5}+...+\frac{100}{98.99.100}-\frac{99}{98.99.100}\right)\)
C = \(\frac{3}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
C = \(\frac{3}{2}.\left(\frac{1}{2.3}-\frac{1}{99.100}\right)\)
C = \(\frac{3}{2}.\frac{1649}{9900}\)
C = \(\frac{1649}{6600}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}.\left(\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right)=\frac{1}{2}.\frac{22}{45}=\frac{11}{45}\)
P = 2 x 3 + 3 x 4 + ...+ 99 x 100
=> 3 x P = 2 x 3 x 3 + 3 x 4 x 3 + ....+ 99 x 100 x 3
3 x P = 2 x 3 x ( 4-1) + 3 x 4 x (5-2) + ...+ 99 x 100 x ( 101 -98)
3 x P = 2 x 3 x 4 - 1 x 2 x 3 + 3 x 4 x 5- 2 x 3 x 4 + ...+ 99 x 100 x 101 - 98 x 99 x 101
3 x P = ( 2 x 3 x 4 + 3 x 4 x 5 + ...+ 99 x 100 x 101) - ( 1 x 2 x 3 + 2 x 3 x 4 + ...+ 98 x 99 x 101)
3 x P = 99 x 100 x 101 - 1 x 2 x 3
\(P=\frac{99x100x101-1x2x3}{3}=333298\)
p=2.3+3.4+4.5+5.6+...+99.100
3p=2.3.3+3.4.3+4.5.3+5.6.3+...+99.100.3
3p=2.3.(4-1)+3.4.(5-2)+4.5.(6-3)+5.6.(7-4)+...+99.100.(101-98)
3p=2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+5.6.7-4.5.6+99.100.101-98.99.100
3p=98.99.100-1.2.3
p=\(\frac{98.99.100-1.2.3}{3}=323398\)
( 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 +1/5.6 ) x 10 - x = 0
= ( 1- 1/2 +1/2 -1/3 +1/3 - 1/4 + 1/4 - 1/5 +1/5 -1/6 ) x 10 - x = 0
= ( 1 - 1/6 ) x 10 - x = 0
= 5/6 x 10 - x =0
= 25/3 - x =0
x = 25/3 - 0
x = 25/3
\(\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}\right)\times10-x=0\)
\(\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\right)\times10-x=0\)
\(\left(\frac{1}{1}-\frac{1}{6}\right)\times10-x=0\)
\(\frac{5}{6}\times10-x=0\)
\(\frac{25}{3}-x=0\)
x =\(\frac{25}{3}-0=\frac{25}{3}\)
\(1,\\ =\dfrac{2-1}{1\times2}+\dfrac{3-2}{2\times3}+\dfrac{4-3}{3\times4}+\dfrac{5-4}{4\times5}+.....+\dfrac{99-98}{98\times99}+\dfrac{100-99}{99\times100}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+....+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}=\dfrac{100-1}{100}=\dfrac{99}{100}\)
\(2,=\dfrac{13-11}{11\times13}+\dfrac{15-13}{13\times15}+....+\dfrac{21-19}{19\times21}+\dfrac{23-21}{21\times23}\\ =\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+....+\dfrac{1}{19}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{23}\\ =\dfrac{1}{11}-\dfrac{1}{23}\\ =\dfrac{23-11}{11\times23}=\dfrac{12}{253}\)
@seven
a: 1/1*2+1/2*3+...+1/99*100
=1-1/2+1/2-1/3+...+1/99-1/100
=1-1/100
=99/100
b: 2/11*13+2/13*15+...+2/21*23
=1/11-1/13+1/13-1/15+...+1/21-1/23
=1/11-1/23
=12/253
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(=\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{10-9}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)
1/2-1/3+1/3-1/4+...+1/9-1/10
=1/2-1/10
=2/5
Chúc bạn học giỏi và thông minh hơn!
\(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{9\times10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
( GẠCH BỎ CÁC PHÂN SỐ GIỐNG NHAU)
\(=\frac{1}{2}-\frac{1}{10}\)
\(=\frac{5}{10}-\frac{1}{10}\)
\(=\frac{4}{10}=\frac{2}{5}\)
CHÚC BẠN HỌC TỐT!!!!!!!!
\(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+.....+\frac{1}{9\times10}\)
Đặt \(A=\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+.....+\frac{1}{9\times10}\)
Nhận xét:
\(\frac{1}{2\times3}=\frac{1}{2}-\frac{1}{3};\frac{1}{3\times4}=\frac{1}{3}-\frac{1}{4}\)
\(\frac{1}{4\times5}=\frac{1}{4}-\frac{1}{5};......;\frac{1}{9\times10}=\frac{1}{9}-\frac{1}{10}\)
Do đó \(A=\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=\frac{1}{2}-\frac{1}{10}\)
\(A=\frac{2}{5}\)