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c) Ta có: \(\dfrac{3}{5}+\dfrac{-5}{20}+\dfrac{30}{75}+\dfrac{-7}{4}\)
\(=\dfrac{3}{5}+\dfrac{2}{5}+\dfrac{-1}{4}+\dfrac{-7}{4}\)
\(=1-2=-1\)
Giải:
a)-1/12+4/3=-1/12+16/12=15/12=5/4
b)(-4/14-3/15)-(1/5-20/35-(-1)).7
=-17/35-22/35.7
=-17/35-22/5
=-171/35
c)3/5+-5/20+30/75+-7/4
=3/5+-1/4+2/5+-7/4
=(3/5+2/5)+(-1/4+-7/4)
=1+-2
=-1
d)5/6.-12/14+7/13
=-5/7+7/13
=-16/91
e)2/-9-5/-36-1/4
=-1/12-1/4
=-1/3
f)2/23+-5/12+7/18+21/23+-7/12
=(2/23+21/23)+(-5/12+-7/12)+7/18
=1+-1+7/18
=7/18
`5`
`a, -7/21 +(1+1/3)`
`=-7/21 + ( 3/3 + 1/3)`
`=-7/21+ 4/3`
`=-7/21+ 28/21`
`= 21/21`
`=1`
`b, 2/15 + ( 5/9 + (-6)/9)`
`= 2/15 + (-1/9)`
`= 1/45`
`c, (9-1/5+3/12) +(-3/4)`
`= ( 45/5-1/5 + 3/12)+(-3/4)`
`= ( 44/5 + 3/12)+(-3/4)`
`= 9,05 +(-0,75)`
`=8,3`
`6`
`x+7/8 =13/12`
`=>x= 13/12 -7/8`
`=>x=5/24`
`-------`
`-(-6)/12 -x=9/48`
`=> 6/12 -x=9/48`
`=>x= 6/12-9/48`
`=>x=5/16`
`---------`
`x+4/6 =5/25 -(-7)/15`
`=>x+4/6 =1/5 + 7/15`
`=> x+ 4/6=10/15`
`=>x=10/15 -4/6`
`=>x=0`
`----------`
`x+4/5 = 6/20 -(-7)/3`
`=>x+4/5 = 6/20 +7/3`
`=>x+4/5 = 79/30`
`=>x=79/30 -4/5`
`=>x= 79/30-24/30`
`=>x= 55/30`
`=>x= 11/6`
\(5)\)
\(A=\dfrac{-7}{21}+\left(1+\dfrac{1}{3}\right)\)
\(A=\dfrac{-7}{21}+\dfrac{4}{3}\)
\(A=\dfrac{-7}{21}+\dfrac{28}{21}\)
\(A=1\)
\(--------------\)
\(B=\dfrac{2}{15}+\left(\dfrac{5}{9}+\dfrac{-6}{9}\right)\)
\(B=\dfrac{2}{15}+\dfrac{-1}{9}\)
\(B=\dfrac{18}{135}+\dfrac{-15}{135}\)
\(B=\dfrac{1}{45}\)
\(------------\)
\(C=9-\dfrac{1}{5}+\dfrac{3}{12}+\dfrac{-3}{4}\)
\(C=\dfrac{44}{5}+\dfrac{3}{12}+\dfrac{-3}{4}\)
\(C=\dfrac{528}{60}+\dfrac{15}{60}+\dfrac{-3}{4}\)
\(C=\dfrac{181}{20}+\dfrac{-3}{4}\)
\(C=\dfrac{181}{20}+\dfrac{-15}{20}\)
\(C=\dfrac{83}{10}\)
\(6)\)
\(a)\) \(x+\dfrac{7}{8}=\dfrac{13}{12}\)
\(x=\dfrac{13}{12}-\dfrac{7}{8}\)
\(x=\dfrac{104}{96}-\dfrac{84}{96}\)
\(x=\dfrac{5}{24}\)
\(b)\) \(\dfrac{-6}{12}-x=\dfrac{9}{48}\)
\(\dfrac{-1}{2}-x=\dfrac{3}{16}\)
\(x=\dfrac{-1}{2}-\dfrac{3}{16}\)
\(x=\dfrac{-8}{16}-\dfrac{3}{16}\)
\(x=\dfrac{-11}{16}\)
\(c)\) \(x+\dfrac{4}{6}=\dfrac{5}{25}-\left(-\dfrac{7}{15}\right)\)
\(x+\dfrac{4}{6}=\dfrac{5}{25}+\dfrac{7}{15}\)
\(x+\dfrac{4}{6}=\dfrac{75}{375}+\dfrac{105}{375}\)
\(x+\dfrac{4}{6}=\dfrac{12}{25}\)
\(x=\dfrac{12}{25}-\dfrac{4}{6}\)
\(x=\dfrac{72}{150}-\dfrac{100}{150}\)
\(x=\dfrac{-14}{75}\)
\(d)\) \(x+\dfrac{4}{5}=\dfrac{6}{20}-\left(-\dfrac{7}{3}\right)\)
\(x+\dfrac{4}{5}=\dfrac{6}{20}+\dfrac{7}{3}\)
\(x+\dfrac{4}{5}=\dfrac{18}{60}+\dfrac{140}{60}\)
\(x+\dfrac{4}{5}=\dfrac{79}{30}\)
\(x=\dfrac{79}{30}-\dfrac{4}{5}\)
\(x=\dfrac{79}{30}-\dfrac{24}{30}\)
\(x=\dfrac{11}{6}\)
1)C= 1/5+1/10+1/20+1/40+...+1/1280
\(=\frac{1}{5}\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\right)\)
Đặt cái trong ngoặc là A ta có:\(2A=2+1+...+\frac{1}{2^7}\)
\(2A-A=\left(2+1+...+\frac{1}{2^7}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2^8}\right)\)
\(A=2-\frac{1}{2^8}\).Thay A vào ta được:\(C=\frac{1}{5}\left(2-\frac{1}{2^8}\right)=\frac{1}{5}\cdot\frac{511}{256}=\frac{511}{1280}\)
2)D= 2/1*3+2/3*5+2/5*10+2/7*9+2/9*11+2/11*18+2/13*15
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{13}-\frac{1}{15}\)
\(=1-\frac{1}{15}\)
\(=\frac{14}{15}\)
3)E= 4/3*7+4/7*11+4/11*15+4/15*19+4/19*23+4/23*27
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}\)
\(=\frac{1}{3}-\frac{1}{27}\)
\(=\frac{8}{27}\)
4)G= 1/2+1/6+1/12+1/20+1/30+1/42+...+1/110
\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\)
\(=1-\frac{1}{11}\)
\(=\frac{10}{11}\)
5)H= 3/1*2+3/2*3+3/3*4+3/4*5+...+3/9*10
\(=3\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=3\left(1-\frac{1}{10}\right)\)
\(=3\times\frac{9}{10}\)
\(=\frac{27}{10}\).Lần sau bạn đăng ít một thôi nhé
a; - \(\dfrac{10}{13}\) + \(\dfrac{5}{17}\) - \(\dfrac{3}{13}\) + \(\dfrac{12}{17}\) - \(\dfrac{11}{20}\)
= - (\(\dfrac{10}{13}\) + \(\dfrac{3}{13}\)) + (\(\dfrac{5}{17}\) + \(\dfrac{12}{17}\)) - \(\dfrac{11}{20}\)
= - 1 + 1 - \(\dfrac{11}{20}\)
= 0 - \(\dfrac{11}{20}\)
= - \(\dfrac{11}{20}\)
b; \(\dfrac{3}{4}\) + \(\dfrac{-5}{6}\) - \(\dfrac{11}{-12}\)
= \(\dfrac{9}{12}\) - \(\dfrac{10}{12}\) + \(\dfrac{11}{12}\)
= \(\dfrac{10}{12}\)
= \(\dfrac{5}{6}\)
c; [13.\(\dfrac{4}{9}\) + 2.\(\dfrac{1}{9}\)] - 3.\(\dfrac{4}{9}\)
= [\(\dfrac{52}{9}\) + \(\dfrac{2}{9}\)] - \(\dfrac{4}{3}\)
= \(\dfrac{54}{9}\) - \(\dfrac{4}{3}\)
= \(\dfrac{14}{3}\)
Bài 1:
\(A=\dfrac{-1}{3}+1+\dfrac{1}{3}=1\)
\(B=\dfrac{2}{15}+\dfrac{5}{9}-\dfrac{6}{9}=\dfrac{2}{15}-\dfrac{1}{9}=\dfrac{18-15}{135}=\dfrac{3}{135}=\dfrac{1}{45}\)
\(C=\dfrac{-1}{5}+\dfrac{1}{4}-\dfrac{3}{4}=\dfrac{-1}{5}-\dfrac{1}{2}=\dfrac{-7}{10}\)
Bài 2:
a: \(=\dfrac{1}{5}+\dfrac{1}{2}+\dfrac{2}{5}-\dfrac{3}{5}+\dfrac{2}{21}-\dfrac{10}{21}+\dfrac{3}{20}\)
\(=\left(\dfrac{1}{5}+\dfrac{2}{5}-\dfrac{3}{5}\right)+\left(\dfrac{2}{21}-\dfrac{10}{21}\right)+\left(\dfrac{1}{2}+\dfrac{3}{20}\right)\)
\(=\dfrac{-8}{21}+\dfrac{13}{20}=\dfrac{113}{420}\)
b: \(B=\dfrac{21}{23}-\dfrac{21}{23}+\dfrac{125}{93}-\dfrac{125}{143}=\dfrac{6250}{13299}\)
Bài 3:
\(\dfrac{7}{3}-\dfrac{1}{2}-\left(-\dfrac{3}{70}\right)=\dfrac{7}{3}-\dfrac{1}{2}+\dfrac{3}{70}=\dfrac{490}{210}-\dfrac{105}{210}+\dfrac{9}{210}=\dfrac{394}{210}=\dfrac{197}{105}\)
\(\dfrac{5}{12}-\dfrac{3}{-16}+\dfrac{3}{4}=\dfrac{5}{12}+\dfrac{3}{16}+\dfrac{3}{4}=\dfrac{20}{48}+\dfrac{9}{48}+\dfrac{36}{48}=\dfrac{65}{48}\)
Bài 4:
\(\dfrac{3}{4}-x=1\)
\(\Rightarrow-x=1-\dfrac{3}{4}\)
\(\Rightarrow x=-\dfrac{1}{4}\)
Vậy: \(x=-\dfrac{1}{4}\)
\(x+4=\dfrac{1}{5}\)
\(\Rightarrow x=\dfrac{1}{5}-4\)
\(\Rightarrow x=-\dfrac{19}{5}\)
Vậy: \(x=-\dfrac{19}{5}\)
\(x-\dfrac{1}{5}=2\)
\(\Rightarrow x=2+\dfrac{1}{5}\)
\(\Rightarrow x=\dfrac{11}{5}\)
Vậy: \(x=\dfrac{11}{5}\)
\(x+\dfrac{5}{3}=\dfrac{1}{81}\)
\(\Rightarrow x=\dfrac{1}{81}-\dfrac{5}{3}\)
\(\Rightarrow x=-\dfrac{134}{81}\)
Vậy: \(x=-\dfrac{134}{81}\)
\(B=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+\frac{1}{18\cdot19\cdot20}\)
\(B=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+\frac{2}{18\cdot19\cdot20}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{18\cdot19}-\frac{1}{19\cdot20}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{19\cdot20}\right)\)
\(B=\frac{1}{2}\cdot\frac{189}{380}=\frac{189}{760}\)
\(C=\frac{52}{1\cdot6}+\frac{52}{6\cdot11}+\frac{52}{11\cdot16}+...+\frac{52}{31\cdot36}\)
\(C=\frac{52}{5}\left(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+\frac{5}{11\cdot16}+...+\frac{6}{31\cdot36}\right)\)
\(C=\frac{52}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{31}-\frac{1}{36}\right)\)
\(C=\frac{52}{5}\cdot\left(1-\frac{1}{36}\right)\)
\(C=\frac{91}{9}\)