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\(=\left(\dfrac{4}{4}-\dfrac{1}{4}\right)\left(\dfrac{9}{9}-\dfrac{1}{9}\right)\left(\dfrac{16}{16}-\dfrac{1}{16}\right)...\left(\dfrac{10000}{10000}-\dfrac{1}{10000}\right)\)
\(=\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}....\cdot\dfrac{9999}{10000}\)
\(=\dfrac{3.8.15.....9999}{4.9.16.....10000}=\dfrac{\left(1.3\right)\left(2.4\right)\left(3.5\right)....\left(99.101\right)}{\left(2.2\right)\left(3.3\right)\left(4.4\right).....\left(100.100\right)}\)
\(=\dfrac{\left(1.2.3...99\right)\left(3.4.5....101\right)}{\left(2.3.4...100\right)\left(2.3.4...101\right)}=\dfrac{101.1}{100.2}=\dfrac{101}{200}\)
`A = 3/4 xx 8/9 xx ... xx 99/100`
`= (1xx3)/(2xx2) xx (2xx4)/(3xx3) xx ... xx (9xx11)/(10xx10)`
`= (1xx2xx3xx ... xx 9)/(2xx3xx...xx10) xx (3xx4xx5xx...xx 11)/(2xx3xx4xx...xx 10)`
`= 1/10 xx 11`
`= 11/10`.
Ta có: `11/10 > 1`
`11/19 < 1`.
`=> A > 11/19`.
a: =>4y+15/16=1
=>4y=1/16
hay y=1/64
b: =>10y+1023/1024=1
=>10y=1/1024
hay y=1/10240
`1//([-1]/2)^2 . |+8|-(-1/2)^3:|-1/16|=1/4 .8+1/8 .16=2+2=4`
`2//|-0,25|-(-3/2)^2:1/4+3/4 .2017^0=0,25-2,25.4+0,75.1=0,25-9+0,75=-8,75+0,75-8`
`3//|2/3-5/6|.(3,6:2 2/5)^3=|-1/6|.(3/2)^3=1/6 . 27/8=9/16`
`4//|(-0,5)^2+7/2|.10-(29/30-7/15):(-2017/2018)^0=|1/4+7/2|.10-1/2:1=|15/4|.10-1/2=15/4 .10-1/2=75/2-1/2=37`
`5// 8/3+(3-1/2)^2-|[-7]/3|=8/3+(5/2)^2-7/3=8/3+25/4-7/3=107/12-7/3=79/12`
a) Ta có: \(2\dfrac{3}{3}\cdot4\cdot\left(-0.4\right)+1\dfrac{3}{5}\cdot1.75+\left(-7.2\right):\dfrac{9}{11}\)
\(=-4.8+\dfrac{8}{5}\cdot\dfrac{7}{4}-\dfrac{36}{5}\cdot\dfrac{11}{9}\)
\(=\dfrac{-24}{5}+\dfrac{14}{5}-\dfrac{44}{5}\)
\(=\dfrac{-54}{5}\)
b) Ta có: \(\left(\dfrac{1}{24}-\dfrac{5}{16}\right):\dfrac{-3}{8}+1^{10}\cdot\left(-5\right)^0\)
\(=\left(\dfrac{2}{48}-\dfrac{15}{48}\right)\cdot\dfrac{8}{-3}+1\cdot1\)
\(=\dfrac{-13}{48}\cdot\dfrac{-8}{3}+1\)
\(=\dfrac{13}{18}+\dfrac{18}{18}=\dfrac{31}{18}\)
A = (\(\dfrac{5}{6}\) - \(\dfrac{4}{5}\)) . 1\(\dfrac{1}{5}\) + \(\dfrac{3}{16}\) : (\(\dfrac{-1}{2}\))3
A = \(\dfrac{1}{30}\) . \(\dfrac{6}{5}\) + \(\dfrac{3}{16}\) : \(\dfrac{-1}{8}\)
A = \(\dfrac{1}{25}\) + \(\dfrac{3}{16}\) . \(\dfrac{-8}{1}\)
A = \(\dfrac{1}{25}\) + \(\dfrac{-3}{2}\)
A = \(\dfrac{-73}{50}\)
B = \(\dfrac{4}{17}\) . (7\(\dfrac{3}{4}\) - 6\(\dfrac{1}{3}\)) + (5\(\dfrac{3}{4}\) - 6.95) : (-1\(\dfrac{3}{5}\))
B = \(\dfrac{4}{17}\) . \(\dfrac{17}{12}\) + (\(\dfrac{23}{4}\) - \(\dfrac{139}{20}\)) : \(\dfrac{-8}{5}\)
B = \(\dfrac{1}{3}\) + \(\dfrac{-6}{5}\) . \(\dfrac{-5}{8}\)
B = \(\dfrac{13}{12}\)
a: =>1/3x+2/5x-2/5=0
=>11/15x-2/5=0
=>11/15x=2/5
=>x=2/5:11/15=2/5*15/11=30/55=6/11
b: =>-5x-1-1/2x+1/3=x
=>-11/2x-2/3-x=0
=>-13/2x=2/3
=>x=-2/3:13/2=-2/3*2/13=-4/39
c: (x+1/2)(2/3-2x)=0
=>x+1/2=0 hoặc 2/3-2x=0
=>x=1/3 hoặc x=-1/2
d: 9(3x+1)^2=16
=>(3x+1)^2=16/9
=>3x+1=4/3 hoặc 3x+1=-4/3
=>3x=1/3 hoặc 3x=-7/3
=>x=1/9 hoặc x=-7/9
\(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right).......\left(1-\dfrac{1}{10}\right)\)
\(=\left(\dfrac{2}{2}-\dfrac{1}{2}\right)\left(\dfrac{3}{3}-\dfrac{1}{3}\right).........\left(\dfrac{10}{10}-\dfrac{1}{10}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}......\dfrac{9}{10}\)
\(=\dfrac{1}{10}\)
Ta có :
\(A=\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)\left(1-\dfrac{1}{25}\right)...............\left(1-\dfrac{1}{361}\right)\left(1-\dfrac{1}{400}\right)\)
\(\Rightarrow A=\left(\dfrac{9}{9}-\dfrac{1}{9}\right)\left(\dfrac{16}{16}-\dfrac{1}{16}\right)\left(\dfrac{25}{25}-\dfrac{1}{25}\right).............\left(\dfrac{361}{361}-\dfrac{1}{361}\right)\left(\dfrac{400}{400}-\dfrac{1}{400}\right)\)\(\Rightarrow A=\dfrac{8}{9}.\dfrac{15}{16}.\dfrac{24}{25}................\dfrac{360}{361}.\dfrac{399}{400}\)
\(\Rightarrow A=\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}.\dfrac{4.6}{5^2}...............\dfrac{18.20}{19^2}.\dfrac{19.21}{20^2}\)
\(\Rightarrow A=\dfrac{\left(2.3.4......19\right)\left(4.5.6......21\right)}{\left(3.4.5.....20\right)\left(3.4.5...20\right)}=\dfrac{2.21}{3.20}=\dfrac{7}{10}\)
Sửa đề: \(B=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)\left(\dfrac{1}{16}-1\right)\cdot...\cdot\left(\dfrac{1}{400}-1\right)\)
Ta có: \(B=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)\left(\dfrac{1}{16}-1\right)\cdot...\cdot\left(\dfrac{1}{400}-1\right)\)
\(=\dfrac{-3}{4}\cdot\dfrac{-8}{9}\cdot\dfrac{-15}{16}\cdot...\cdot\dfrac{-399}{400}\)
\(=\dfrac{-3\cdot8\cdot15\cdot...\cdot399}{4\cdot9\cdot16\cdot...\cdot400}\)
\(=\dfrac{-3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot19\cdot21}{2^2\cdot3^2\cdot4^2\cdot...\cdot20^2}\)
\(=\dfrac{-2\cdot3\cdot4\cdot...\cdot19}{2\cdot3\cdot4\cdot...\cdot20}\cdot\dfrac{3\cdot4\cdot5\cdot...\cdot21}{2\cdot3\cdot4\cdot20}\)
\(=\dfrac{-1}{20}\cdot\dfrac{21}{2}\)
\(=\dfrac{-21}{40}\)