Tính tổng sau:
A = \(\dfrac{4}{2.4}+\dfrac{4}{4.6}+\dfrac{4}{6.8}+...+\dfrac{4}{2008.2010}\)
help me!
Tính nhanh: F= \(\dfrac{4}{2.4}+\dfrac{4}{4.6}+\dfrac{4}{6.8}+...+\dfrac{4}{2008.2010}\)
Cảm ơn!
Ta có: \(F=\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{2008\cdot2010}\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)
\(=2\cdot\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)\)
\(=2\cdot\dfrac{502}{1005}=\dfrac{1004}{1005}\)
\(F=\dfrac{4}{2.4}+\dfrac{4}{4.6}+\dfrac{4}{6.8}+...+\dfrac{4}{2008.2010}\)
\(F=2.\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{2008.2010}\right)\)
\(F=2.\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)
\(F=2.\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)\)
\(F=1-\dfrac{1}{1005}=\dfrac{1004}{1005}\)
=2(2/2.4+2/4.6+......+22/008.2010)
=2(12−12010)
tính
F = \(\dfrac{4}{2.4}+\dfrac{4}{4.6}+\dfrac{4}{6.8}+...+\dfrac{4}{2008.2010}\)
Ta có :
\(F=\dfrac{4}{2.4}+\dfrac{4}{4.6}+..................+\dfrac{4}{2008.2010}\)
\(\Rightarrow F=2\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+.............+\dfrac{2}{2008.2010}\right)\)
\(\Rightarrow F=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+..............+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)
\(\Rightarrow F=2\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)\)
\(\Rightarrow F=2.\dfrac{502}{1005}=\dfrac{1004}{1005}\)
\(F=\dfrac{4}{2.4}+\dfrac{4}{4.6}+\dfrac{4}{6.8}+......+\dfrac{4}{2008.2010}\)
\(F=\dfrac{4}{2}\left(\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+.....+\dfrac{1}{2008.2010}\right)\)
\(F=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+.....+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)\(F=2\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)\)\(F=2.\dfrac{502}{1005}\)
\(F=\dfrac{1004}{1005}\)
tính nhanh nếu có thể:
a)\(\dfrac{4}{2.4}+\dfrac{4}{4.6}+\dfrac{4}{6.8}+.....+\dfrac{4}{2008.2010}\)
b)\(\dfrac{\dfrac{3}{41}-\dfrac{12}{47}+\dfrac{27}{53}}{\dfrac{4}{41}-\dfrac{16}{47}+\dfrac{36}{53}}\)
a)\(\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{2008\cdot2010}\)
\(=2\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{2008\cdot2010}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)=2\cdot\dfrac{502}{1005}=\dfrac{1004}{1005}\)
b)\(\dfrac{\dfrac{3}{41}-\dfrac{12}{47}+\dfrac{27}{53}}{\dfrac{4}{41}-\dfrac{16}{47}+\dfrac{36}{53}}=\dfrac{3\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}{4\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}=\dfrac{3}{4}\)
a) gọi biểu thức đó là A
Ta có công thức \(\dfrac{a}{b.c}=\dfrac{a}{c-b}.\left(\dfrac{1}{b}-\dfrac{1}{c}\right)\)
Dựa vào công thức trên, ta có
\(A=\dfrac{4}{2}.\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+....+\dfrac{1}{2008}-\dfrac{1}{2009}\right)\)
\(A=\dfrac{4}{2}.\left(\dfrac{1}{2}-\dfrac{1}{2009}\right)\)
\(A=2.\left(\dfrac{2007}{4018}\right)=\dfrac{2007}{2009}\)
b) dễ quá bạn tự làm. (không phải mink không biết làm đâu nha)
Cho A= \(\dfrac{4}{2.4}\)+\(\dfrac{4}{4.6}\)+\(\dfrac{4}{6.8}\)+...+\(\dfrac{4}{46.48}\)+\(\dfrac{4}{48.50}\)
\(A=2\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{48.50}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{48}-\dfrac{1}{50}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{50}\right)\)
\(=2\times\dfrac{12}{25}=\dfrac{24}{25}\)
\(=>A=4.\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{46}-\dfrac{1}{48}+\dfrac{1}{48}-\dfrac{1}{50}\right)\)
\(A=4.\left(\dfrac{1}{2}-\dfrac{1}{50}\right)=4.\left(\dfrac{25}{50}-\dfrac{1}{50}\right)=\dfrac{4.24}{50}=\dfrac{48}{25}\)
\(\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}.\dfrac{5^2}{4.6}\)
Help me, các chế ơi
\(\dfrac{4}{3}\times\dfrac{9}{8}\times\dfrac{16}{15}\times\dfrac{25}{24}=\dfrac{5}{3}\)
`(2^2)/(1 . 3) . (3^2)/(2 . 4) . (4^2)/(3 . 5) . (5^2)/(4 . 6)`
`= 4/3 . 9/8 . 16/15 . 25/24 = 5/3`
\(l,\dfrac{\dfrac{3}{41}-\dfrac{12}{47}+\dfrac{27}{53}}{\dfrac{4}{41}-\dfrac{16}{47}+\dfrac{36}{53}}\)
\(m,\left(3-2\dfrac{1}{3}+\dfrac{1}{4}\right):\left(4-5\dfrac{1}{6}+2\dfrac{1}{4}\right)\)
\(n,F=\dfrac{4}{2.4}+\dfrac{4}{4.6}+\dfrac{4}{6.8}+...+\dfrac{4}{2008.2010}\)
\(p,F=\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+...+\dfrac{1}{990}\)
p: \(F=\dfrac{1}{3}\left(\dfrac{3}{3\cdot6}+\dfrac{3}{6\cdot9}+\dfrac{3}{9\cdot12}+...+\dfrac{3}{30\cdot33}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{30}-\dfrac{1}{33}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{10}{33}=\dfrac{10}{99}\)
n: \(F=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)
\(=2\cdot\dfrac{502}{1005}=\dfrac{1004}{1005}\)
m: \(=\left(3-\dfrac{7}{3}+\dfrac{1}{4}\right):\left(4-\dfrac{31}{6}+\dfrac{9}{4}\right)\)
\(=\dfrac{36-28+3}{12}:\dfrac{48-62+27}{12}\)
\(=\dfrac{11}{13}\)
Tính :
a. F = \(\dfrac{4}{2.4}+\dfrac{4}{4.6}+\dfrac{4}{6.8}+....+\dfrac{4}{2008.2010}\)
b. C = \(\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+....+\dfrac{1}{990}\)
c. T = \(1+\dfrac{1}{2}.\left(1+2\right)+\dfrac{1}{3}.\left(1+2+3\right)+....+\dfrac{1}{16}.\left(1+2+3+....+16\right)\)
d. H = \(\dfrac{7}{4}.\left(\dfrac{33}{12}+\dfrac{3333}{2020}+\dfrac{333333}{303030}+\dfrac{33333333}{42424242}\right)\)
( Các bạn giúp mk , làm được 1 câu cũng được , làm hết thì càng tốt )
\(b,C=\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+...+\dfrac{1}{990}\\ =\dfrac{1}{3.6}+\dfrac{1}{6.9}+\dfrac{1}{9.12}+...+\dfrac{1}{30.33}\\ =\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{12}+...+\dfrac{1}{30}-\dfrac{1}{33}\\ =\dfrac{1}{3}-\dfrac{1}{33}\\ =\dfrac{11}{33}-\dfrac{1}{33}=\dfrac{10}{33}\)
a.F=\(\dfrac{4}{2.4}\)+\(\dfrac{4}{4.6}\)+\(\dfrac{4}{6.8}\)+...+\(\dfrac{4}{2008.2010}\)
F=\(\dfrac{2.2}{2.4}\)+\(\dfrac{2.2}{4.6}\)+\(\dfrac{2.2}{6.8}\)+...+\(\dfrac{2.2}{2008.2010}\)
F=2.(\(\dfrac{2}{2.4}\)+\(\dfrac{2}{4.6}\)+\(\dfrac{2}{6.8}\)+...+\(\dfrac{2}{2008.2010}\))
F=2.(\(\dfrac{1}{2}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{8}\)+...+\(\dfrac{1}{2008}\)-\(\dfrac{1}{2010}\))
F=2.(\(\dfrac{1}{2}\)-\(\dfrac{1}{2010}\))
F=\(\dfrac{1004}{1005}\)
b, C=\(\dfrac{1}{18}\)+\(\dfrac{1}{54}\)+....+\(\dfrac{1}{990}\)
\(\Rightarrow\)C=\(\dfrac{1}{3.6}\)+\(\dfrac{1}{6.9}\)+...+\(\dfrac{1}{30.33}\)
=>3C=3( \(\dfrac{1}{3.6}\)+\(\dfrac{1}{6.9}\)+...+\(\dfrac{1}{30.33}\))
=>3C=\(\dfrac{3}{3.6}\)+\(\dfrac{3}{6.9}\)+....+\(\dfrac{3}{30.33}\)
=> 3C=\(\dfrac{1}{3}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{9}\)+...+\(\dfrac{1}{30}\)-\(\dfrac{1}{33}\)
=> 3C= \(\dfrac{1}{3}\)-\(\dfrac{1}{33}\)
=>3C=\(\dfrac{10}{33}\)
=> C=\(\dfrac{10}{33}\):3
=> C=\(\dfrac{10}{99}\)
a, F=4/2.4+4/4/6+4/6.8+......+4/2008.2010
=> F= 4/2.(2/2.4+2/4.6+2/6.8+......+2/2008/2010
=> F= 4/2. ( 1/2-1/4+1/4-1/6+1/6-1/8+......+1/2008-1/2010
=> F=4/2.( 1/2-1/2010)
=> F= 4/2. 502/1005
=> F= \(\dfrac{1004}{1005}\)
Tính A=4/2.4+4/4.6+4/6.8+....+4/2008.2010
A=4/2.4+4/4.6+4/6.8+...+4/2008.2010
=2.(2/2.4+2/4.6+2/6.8+...+2/2008.2010)
=2.(1/2-1/4+1/4-1/6+1/6-1/8+...+1/2008-1/2010)
=2.(1/2-1/2010)
=2.502/1005
=1004/1005
Vậy A=1004/1005
100% giải đúng đầu tiên:
Ta có: \(A=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(=2.\frac{2}{2.4}+2.\frac{2}{4.6}+2.\frac{2}{6.8}+...+2.\frac{2}{2008.2010}\)
\(=2\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+..+\frac{2}{2008.2010}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(=2.\frac{1}{2}-2.\frac{1}{2010}\)
\(=1-\frac{1}{1005}=\frac{1004}{1005}\)
Tính giá trị của biểu thức sau
F= 4/2.4+4/4.6+4/6.8 +...+4/2008.2010
F=2\ 2/2.4+2/4.6+2/6.8+.....+2/2008.2010 \
=2 \ 1/2-1/4+1/4-1/6+1/6-1/8+.....+1/2008-1/2010 \
=2 \ 1/2-1/2010 \ =2 \ 502/1005 \ =1004/1005
chú ý : \ là ngoặc