\(\dfrac{3}{2}\) + \(\dfrac{3}{8}\) + \(\dfrac{3}{24}\) + ...........+ \(\dfrac{3}{9800}\) + \(\dfrac{3}{10200}\) = ?
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Y
3 tháng 3 2019
\(A=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\)
\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A=1-\dfrac{1}{100}=\dfrac{99}{100}\)
\(B=\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+..+\dfrac{1}{195}\) ( là 195 ms đúng ! )
\(B=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{13\cdot15}\)
\(B=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{13}-\dfrac{1}{15}\right)\)
\(B=\dfrac{1}{2}\left(1-\dfrac{1}{15}\right)=\dfrac{1}{2}\cdot\dfrac{14}{15}=\dfrac{7}{15}\)
\(C=\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+...+\dfrac{1}{98\cdot100}\)
Rồi làm tương tự cân b nha!
\(D=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{32}+\dfrac{1}{32}-\dfrac{1}{57}\)
\(+\dfrac{1}{57}-\dfrac{1}{87}\)
\(D=\dfrac{1}{3}-\dfrac{1}{87}=\dfrac{28}{87}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
19 tháng 3
A = \(\dfrac{3}{4}\).\(\dfrac{8}{9}\).\(\dfrac{15}{16}.\)\(\dfrac{24}{25}\)...\(\dfrac{9800}{9801}\)
A = \(\dfrac{1.3}{2.2}\).\(\dfrac{2.4}{3.3}\).\(\dfrac{3.5}{4.4}\)...\(\dfrac{98.100}{99.99}\)
A = \(\dfrac{1}{2}.\dfrac{100}{99}\)
A = \(\dfrac{50}{99}\)
B = \(\dfrac{1.2+2.3+3.4+...+98.99}{98.99.100}\)
Đặt tử số là C Thì
C = 1.2 + 2.3 + 3.4 +...+ 98.99
C = \(\dfrac{1}{3}\).(1.2.3 + 2.3.3 + 3.4.3 + ...+ 98.99.3)
C = \(\dfrac{1}{3}\).[1.2.3 + 2.3.(4-1) + 3.4.(5-2) +...+ 98.99.(100-97)]
C = \(\dfrac{1}{3}\).[1.2.3 -1.2.3+2.3.4- 2.3.4 + 2.4.5 - .... - 97.98.99 + 98.99.100]
C = \(\dfrac{1}{3}\).98.99.100
B = \(\dfrac{\dfrac{1}{3}.98.99.100}{98.99.100}\)
B = \(\dfrac{1}{3}\) = \(\dfrac{33}{99}\) < \(\dfrac{50}{99}\) = A
Vậy B < A
29 tháng 6 2023
a: =>6/x=x/24
=>x^2=144
=>x=12 hoặc x=-12
b: =>x(1-7/12+3/8)=5/24
=>x*19/24=5/24
=>x=5/24:19/24=5/19
c: =>(x-1/3)^2=1+3/4+1/2=9/4
=>x-1/3=3/2 hoặc x-1/3=-3/2
=>x=11/6 hoặc x=-7/6
d: =>(x-3)^2=16
=>x-3=4 hoặc x-3=-4
=>x=-1 hoặc x=7
e: =>9/x=-1/3
=>x=-27
f: =>x-1/2=0 hoặc -x/2-3=0
=>x=1/2 hoặc x=-6
14 tháng 3 2017
\(6.-\dfrac{5}{18}=-\dfrac{5}{3}\)
\(-\dfrac{13}{24}+-\dfrac{5}{21}.\dfrac{7}{25}=-\dfrac{13}{24}+-\dfrac{1}{15}=-\dfrac{73}{120}\)
\(-\dfrac{5}{13}-\dfrac{3}{26}.-\dfrac{2}{3}=-\dfrac{5}{13}-\left(-\dfrac{1}{13}\right)=-\dfrac{4}{13}\)
\(\left(\dfrac{2}{3}+-\dfrac{7}{6}\right).\left(\dfrac{3}{8}+-\dfrac{3}{24}\right)=-\dfrac{1}{2}.\dfrac{1}{4}=\dfrac{1}{8}\)
14 tháng 3 2017
Kết quả lần lượt của các phép tính là :
\(\dfrac{-5}{3}\) , \(\dfrac{-73}{120}\) , \(\dfrac{-4}{13}\) , \(\dfrac{-1}{8}\)
12 tháng 7 2023
a: 4/9+3/7=28/63+27/63=55/63
3/4+7/24=18/24+7/24=25/24
1/3+2/9+4/27=9/27+6/27+4/27=19/27
b: 5/6-3/8=20/24-9/24=11/24
7/15-11/30=14/30-11/30=3/30=1/10
2/3+1/6-7/12
=8/12+2/12-7/12
=3/12=1/4
c: 18/25*15/6=15/25*18/6=3*3/5=9/5
30/49:6/7=30/49*7/6=210/294=5/7
1/2*3/4:6/5=3/8*5/6=15/48=5/16
d: 8*3/5:12/5=24/5*5/12=2
4:9/5:10/3=4*5/9*3/10=2/3
26 tháng 9 2021
a)\(\sqrt{\dfrac{2}{2-\sqrt{3}}}=\sqrt{\dfrac{2\left(2+\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}}\)\(=\sqrt{2\left(2+\sqrt{3}\right)}=\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
b)\(\sqrt{\dfrac{2}{3}}-\sqrt{24}+2\sqrt{\dfrac{3}{8}}+\dfrac{1}{6}=\dfrac{\sqrt{6}}{3}-\sqrt{2^2.6}+\dfrac{2\sqrt{24}}{8}+\dfrac{1}{6}\)
\(=\dfrac{\sqrt{6}}{3}-2\sqrt{6}+\dfrac{\sqrt{2^2.6}}{4}+\dfrac{1}{6}=\dfrac{-5\sqrt{6}}{3}+\dfrac{2\sqrt{6}}{4}+\dfrac{1}{6}\)
\(=\dfrac{-7\sqrt{6}}{6}+\dfrac{1}{6}\)
26 tháng 9 2021
Nhầm xíu
\(\sqrt{\dfrac{2}{2-\sqrt{3}}}-\sqrt{\dfrac{2}{2+\sqrt{3}}}=\sqrt{\dfrac{2\left(2+\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}}-\sqrt{\dfrac{2\left(2-\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}}\)
\(=\sqrt{2\left(2+\sqrt{3}\right)}-\sqrt{2\left(2-\sqrt{3}\right)}=\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\sqrt{3}+1-\left|\sqrt{3}-1\right|=\sqrt{3}+1-\left(\sqrt{3}-1\right)=2\)
Ta có :
\(\dfrac{3}{2}\) = \(\dfrac{3}{1.2}\) = 3 x \(\dfrac{1}{1.2}\) = 3 x ( 1 - \(\dfrac{1}{2}\) ) : 2 ;
\(\dfrac{3}{8}\) = \(\dfrac{3}{2.4}\) = \(\)3 x \(\dfrac{1}{2.4}\) = 3 x ( \(\dfrac{1}{2}\) - \(\dfrac{1}{4}\) ) : 2 ;
\(\dfrac{3}{24}\) = \(\dfrac{3}{4.6}\) = 3 x \(\dfrac{1}{4.6}\) = 3. ( \(\dfrac{1}{4}\) - \(\dfrac{1}{6}\) ) : 2 ;
\(\dfrac{3}{9800}\) = \(\dfrac{3}{98.100}\) = 3 x \(\dfrac{1}{98.100}\) = 3 x ( \(\dfrac{1}{98}\) - \(\dfrac{1}{100}\) ) : 2 ;
\(\dfrac{3}{10200}\) = \(\dfrac{3}{100.102}\) = 3 x \(\dfrac{1}{100.102}\)= 3 x ( \(\dfrac{1}{100}\) - \(\dfrac{1}{102}\) )
Vậy : \(\dfrac{3}{2}\) + \(\dfrac{3}{8}\) + \(\dfrac{3}{24}\) + ........ + \(\dfrac{3}{9800}\) + \(\dfrac{3}{10200}\)
= \(\dfrac{3}{1.2}\) + \(\dfrac{3}{2.4}\) + \(\dfrac{3}{4.6}\) +........+ \(\dfrac{3}{98.100}\) + \(\dfrac{3}{100.102}\)
= 3 x ( \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.4}\) + \(\dfrac{1}{4.6}\) + .........+ \(\dfrac{1}{98.100}\) + \(\dfrac{1}{100.102}\) )
= 3 x ( \(\dfrac{2}{1.2}\) + \(\dfrac{2}{2.4}\) + \(\dfrac{2}{4.6}\) +.................+ \(\dfrac{2}{98.100}\) + \(\dfrac{2}{100.102}\) ) : 2
= 3 x ( 1 - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{6}\) +.......+\(\dfrac{1}{98}\) - \(\dfrac{1}{100}\) + \(\dfrac{1}{100}\) - \(\dfrac{1}{102}\) ) : 2
= 3 x ( 1 - \(\dfrac{1}{102}\) ) : 2 = 3 x \(\dfrac{101}{102}\) : 2
= \(\dfrac{101}{68}\)