tính nhanh
\(\dfrac{4}{1\times3\times5}+\dfrac{4}{3\times5\times7}+...+\dfrac{4}{19\times21\times23}+\dfrac{6}{21\times23\times25}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(1,\\ =\dfrac{2-1}{1\times2}+\dfrac{3-2}{2\times3}+\dfrac{4-3}{3\times4}+\dfrac{5-4}{4\times5}+.....+\dfrac{99-98}{98\times99}+\dfrac{100-99}{99\times100}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+....+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}=\dfrac{100-1}{100}=\dfrac{99}{100}\)
\(2,=\dfrac{13-11}{11\times13}+\dfrac{15-13}{13\times15}+....+\dfrac{21-19}{19\times21}+\dfrac{23-21}{21\times23}\\ =\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+....+\dfrac{1}{19}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{23}\\ =\dfrac{1}{11}-\dfrac{1}{23}\\ =\dfrac{23-11}{11\times23}=\dfrac{12}{253}\)
@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
a) \(\dfrac{30\times25\times7\times8}{75\times8\times12\times14}=\dfrac{3\times2\times5\times25\times7\times8}{25\times3\times8\times3\times4\times2\times7}=\dfrac{5}{3\times4}=\dfrac{5}{12}\)
b) \(\dfrac{8\times3\times4}{16\times3}=\dfrac{8\times3\times2\times2}{8\times2\times3}=2\)
c) \(\dfrac{4\times5\times6}{3\times10\times8}=\dfrac{4\times5\times3\times2}{3\times5\times2\times4\times2}=\dfrac{1}{2}\)
Gọi A = 1.3+3.5+5.7+...+21.23
=> A = 1.(1+2)+3.(3+2)+5.(5+2)+...+21.(21+2)
=> A = 12+1.2+32+2.3+52+2.5+...+212+2.21
=> A = 12+32+52+...+212+(1.2+3.2+5.2+...+2.21)
Gọi B = 12+32+52+...+212
=> B = (21.22.23)/3
Gọi C = 1.2+2.3+5.2+...+2.21
=> C = 2(1+3+5+...+21)
=> C = 2{(21+1).[(21-1):2+1]}/2
=> C = 22x11=242
Vậy A = (21.22.23)/3+242
\(B=\dfrac{40404}{70707}+\dfrac{244\times395-151}{244+395\times243}+\dfrac{1\times3\times5+2\times6\times10+4\times12\times20+7\times21\times35}{1\times5\times7+2\times10\times14+4\times20\times28+7\times35\times49}\\ =\dfrac{4}{7}+\dfrac{243\times395+395-151}{244+395\times243}+\dfrac{1\times3\times5\left(1+2+4+7\right)}{1\times5\times7\left(1+2+4+7\right)}\\ =\dfrac{4}{7}+\dfrac{243\times395+244}{244+395\times243}+\dfrac{3}{7}\\ =\left(\dfrac{4}{7}+\dfrac{3}{7}\right)+1\\ =1+1=2\)
Có: A=\(\dfrac{3}{1.5}+\dfrac{3}{5.10}+...+\dfrac{3}{100.105}\)
=> A=\(3.\dfrac{5}{5}\left(\dfrac{1}{1.5}+\dfrac{1}{5.10}+...+\dfrac{1}{100.105}\right)\)
=> A= \(3.\dfrac{1}{5}\left(\dfrac{5}{1.5}+\dfrac{5}{5.10}+...+\dfrac{5}{100.105}\right)\)
=> A=\(\dfrac{3}{5}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{105}\right)\)
=> A= \(\dfrac{3}{5}\left(1-\dfrac{1}{105}\right)\)=\(\dfrac{3}{5}.\dfrac{104}{105}=\dfrac{312}{525}\)
A =\(\dfrac{4^2}{3\times5}\) \(\times\)\(\dfrac{5^2}{4\times6}\) \(\times\) \(\dfrac{6^2}{5\times7}\) \(\times\) \(\dfrac{7^2}{6\times8}\)
A = \(\dfrac{4\times4\times5^2\times6^2\times7\times7}{3\times4\times5^2\times6^2\times7\times8}\)
A = \(\dfrac{4}{3}\) \(\times\) \(\dfrac{7}{8}\)
A = \(\dfrac{7}{6}\)
Ta có:
\(A=\frac{3}{1\cdot5}+\frac{3}{5\cdot10}+...+\frac{3}{100\cdot105}\)
\(=\frac{3}{5}\cdot\left(\frac{5}{1\cdot5}+\frac{5}{5\cdot10}+...+\frac{5}{100\cdot105}\right)\)
\(=\frac{3}{5}\cdot\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{105}\right)\)
\(=\frac{3}{5}\left(1-\frac{1}{105}\right)=\frac{3}{5}\cdot\frac{104}{105}=\frac{312}{525}\)
1 x 3 x 5 + 2 x 6 x 10 + 4 x 10 x 12 + 7 x 21 x 35 / 1 x 5 x 7 + 2 x 10 x 14 + 4 x 20 x 28 + 7 x 35 x 49
= 3 / 7 + 6 / 14 + 10 x 2 x 6 / 10 x 2 x 28 + 21 / 49
= 3 / 7 + 6 / 14 + 6 / 28 + 21 / 49
= 3 / 7 + 6 / 14 + 3 / 14 + 3 / 7
= ( 3 / 7 + 3 / 7 ) + ( 6 / 14 + 3 / 14 )
= 6 / 7 + 9 14
= 12 / 14 + 9 / 14
= 21 / 14
nho hem
lm dau tien lun do
dung 100% nha
Lời giải:
Gọi tổng trên là $S$
$S=\frac{5-1}{1.3.5}+\frac{7-3}{3.5.7}+...+\frac{23-19}{19.21.23}+\frac{25-21}{21.23.25}$
$=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-....+\frac{1}{19.21}-\frac{1}{21.23}+\frac{1}{21.23}-\frac{1}{23.25}$
$=\frac{1}{1.3}-\frac{1}{23.25}=\frac{572}{1725}$