a, \(\frac{4}{20}+\frac{16}{42}+\frac{6}{15}+\frac{-3}{5}+\frac{-10}{21}+\frac{3}{20}\)

\(=\left(\frac{4}{20}+\frac{-3}{5}+\frac{6}{15}+\frac{3}{20}\right)\)\(+\left(\frac{16}{42}+\frac{2}{21}+\frac{-10}{21}\right)\)

\(=\left(\frac{4}{20}+\frac{-12}{20}+\frac{8}{20}+\frac{3}{20}\right)\)\(+\left(\frac{8}{21}+\frac{2}{21}+\frac{-10}{21}\right)\)

\(=\frac{3}{20}\)

b,\(\frac{42}{46}+\frac{250}{186}+\frac{-2121}{2323}+\frac{-125125}{143143}\)

\(=\frac{21}{23}+\frac{125}{93}+\frac{-21}{23}+\frac{-125}{143}\)

\(=\left(\frac{21}{23}+\frac{-21}{23}\right)+\left(\frac{125}{93}+\frac{-125}{143}\right)\)

\(=\frac{6250}{13299}\)

a: \(C=\dfrac{-1}{2}:\dfrac{-2}{3}:\dfrac{-3}{4}\)

\(=\dfrac{-1}{2}\cdot\dfrac{3}{-2}\cdot\dfrac{4}{-3}\)

\(=-\dfrac{4}{4}=-1\)

b: \(D=\dfrac{-6}{5}:\dfrac{-5}{4}:\dfrac{-4}{3}\)

\(=-\dfrac{6}{5}\cdot\dfrac{4}{5}\cdot\dfrac{3}{4}\)

\(=\dfrac{-18}{25}\)

c: \(\dfrac{4}{20}+\dfrac{16}{42}+\dfrac{6}{15}+\dfrac{-3}{5}+\dfrac{2}{21}+\dfrac{-10}{21}+\dfrac{3}{20}\)

\(=\left(\dfrac{4}{20}+\dfrac{-3}{5}+\dfrac{6}{15}\right)+\left(\dfrac{16}{42}+\dfrac{2}{21}-\dfrac{10}{21}\right)+\dfrac{3}{20}\)

\(=\left(\dfrac{1}{5}-\dfrac{3}{5}+\dfrac{2}{5}\right)+\left(\dfrac{8}{21}+\dfrac{2}{21}-\dfrac{10}{21}\right)+\dfrac{3}{20}\)

\(=\dfrac{3}{20}\)

d: \(\dfrac{42}{46}+\dfrac{250}{186}+\dfrac{-2121}{2323}+\dfrac{-125125}{143143}\)

\(=\dfrac{21}{23}+\dfrac{125}{93}+\dfrac{-21}{21}+\dfrac{-125}{143}\)

\(=\dfrac{125}{93}-\dfrac{125}{143}=\dfrac{6250}{13299}\)

a/ \(=\frac{21}{23}+\frac{125}{143}-\frac{101.21}{101.23}-\frac{1001.125}{1001.143}=0\)

b/ \(=\frac{4}{20}+\frac{8}{21}+\frac{2}{5}-\frac{3}{5}+\frac{2}{21}-\frac{10}{21}+\frac{3}{20}=\frac{7}{20}-\frac{1}{5}=\frac{4}{20}\)

c/ \(\frac{C}{2}=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{420}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{20.21}\)

\(\frac{C}{2}=\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{21-20}{20.21}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{20}-\frac{1}{21}\)

\(\frac{C}{2}=\frac{1}{2}-\frac{1}{21}=\frac{19}{42}\Rightarrow C=\frac{19}{21}\)

`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`

\(5)\)

\(A=\dfrac{-7}{21}+\left(1+\dfrac{1}{3}\right)\)

\(A=\dfrac{-7}{21}+\dfrac{4}{3}\)

\(A=\dfrac{-7}{21}+\dfrac{28}{21}\)

\(A=1\)

\(--------------\)

\(B=\dfrac{2}{15}+\left(\dfrac{5}{9}+\dfrac{-6}{9}\right)\)

\(B=\dfrac{2}{15}+\dfrac{-1}{9}\)

\(B=\dfrac{18}{135}+\dfrac{-15}{135}\)

\(B=\dfrac{1}{45}\)

\(------------\)

\(C=9-\dfrac{1}{5}+\dfrac{3}{12}+\dfrac{-3}{4}\)

\(C=\dfrac{44}{5}+\dfrac{3}{12}+\dfrac{-3}{4}\)

\(C=\dfrac{528}{60}+\dfrac{15}{60}+\dfrac{-3}{4}\)

\(C=\dfrac{181}{20}+\dfrac{-3}{4}\)

\(C=\dfrac{181}{20}+\dfrac{-15}{20}\)

\(C=\dfrac{83}{10}\)

\(6)\)

\(a)\) \(x+\dfrac{7}{8}=\dfrac{13}{12}\)

\(x=\dfrac{13}{12}-\dfrac{7}{8}\)

\(x=\dfrac{104}{96}-\dfrac{84}{96}\)

\(x=\dfrac{5}{24}\)

\(b)\) \(\dfrac{-6}{12}-x=\dfrac{9}{48}\)

\(\dfrac{-1}{2}-x=\dfrac{3}{16}\)

\(x=\dfrac{-1}{2}-\dfrac{3}{16}\)

\(x=\dfrac{-8}{16}-\dfrac{3}{16}\)

\(x=\dfrac{-11}{16}\)

\(c)\) \(x+\dfrac{4}{6}=\dfrac{5}{25}-\left(-\dfrac{7}{15}\right)\)

\(x+\dfrac{4}{6}=\dfrac{5}{25}+\dfrac{7}{15}\)

\(x+\dfrac{4}{6}=\dfrac{75}{375}+\dfrac{105}{375}\)

\(x+\dfrac{4}{6}=\dfrac{12}{25}\)

\(x=\dfrac{12}{25}-\dfrac{4}{6}\)

\(x=\dfrac{72}{150}-\dfrac{100}{150}\)

\(x=\dfrac{-14}{75}\)

\(d)\) \(x+\dfrac{4}{5}=\dfrac{6}{20}-\left(-\dfrac{7}{3}\right)\)

\(x+\dfrac{4}{5}=\dfrac{6}{20}+\dfrac{7}{3}\)

\(x+\dfrac{4}{5}=\dfrac{18}{60}+\dfrac{140}{60}\)

\(x+\dfrac{4}{5}=\dfrac{79}{30}\)

\(x=\dfrac{79}{30}-\dfrac{4}{5}\)

\(x=\dfrac{79}{30}-\dfrac{24}{30}\)

\(x=\dfrac{11}{6}\)

Giải:

\(A=\dfrac{-7}{21}+\left(1+\dfrac{1}{3}\right)\)

\(A=\dfrac{-7}{21}+\dfrac{4}{3}\)

\(A=\dfrac{-7}{21}+\dfrac{28}{21}\)

\(A=\dfrac{21}{21}=1\)

\(B=\dfrac{2}{15}+\left(\dfrac{5}{9}+\dfrac{-6}{9}\right)\)

\(B=\dfrac{2}{15}+\dfrac{-1}{9}\)

\(B=\dfrac{6}{45}+\dfrac{-5}{45}\)

\(B=\dfrac{1}{45}\)

\(C=\left(9-\dfrac{1}{5}+\dfrac{3}{12}\right)+\dfrac{-3}{4}\)

\(C=\left(\dfrac{44}{5}+\dfrac{1}{4}\right)+\dfrac{-3}{4}\)

\(C=\dfrac{181}{20}+\dfrac{-3}{4}\)

\(C=\dfrac{181}{20}+\dfrac{-15}{20}\)

\(C=\dfrac{166}{20}\)

Bài 6:

\(a)x+\dfrac{7}{8}=\dfrac{13}{12}\)

⇔\(x=\dfrac{13}{12}-\dfrac{7}{8}=\dfrac{5}{24}\)

\(b)\dfrac{-6}{12}-x=\dfrac{9}{48}\)

⇔\(x=\dfrac{-6}{12}-\dfrac{9}{48}=\dfrac{-11}{16}\)

\(c)x+\dfrac{4}{6}=\dfrac{5}{25}-\dfrac{-7}{15}\)

\(\Leftrightarrow x+\dfrac{2}{3}=\dfrac{2}{3}\)

\(\Leftrightarrow x=\dfrac{2}{3}-\dfrac{2}{3}=0\)

\(d)x+\dfrac{4}{5}=\dfrac{6}{20}-\dfrac{-7}{3}\)

⇔\(x+\dfrac{4}{5}=\dfrac{79}{30}\)

⇔\(x=\dfrac{79}{30}-\dfrac{4}{5}=\dfrac{11}{6}\)

a) \(\frac{4}{20}+\frac{16}{42}+\frac{6}{15}+\frac{-3}{5}+\frac{2}{21}+\frac{-10}{21}+\frac{3}{20}\)

\(=\frac{1}{5}+\frac{8}{21}+\frac{2}{5}+\frac{-3}{5}+\frac{2}{21}+\frac{-10}{21}+\frac{3}{20}\)

= \(\left(\frac{1}{5}+\frac{2}{5}+\frac{-3}{5}\right)+\left(\frac{8}{21}+\frac{2}{21}+\frac{-10}{21}\right)+\frac{3}{20}\)

\(=\frac{0}{5}+\frac{0}{21}+\frac{3}{20}\)

\(=0+0+\frac{3}{20}\)

\(=\frac{3}{20}\)

 b) \(\frac{42}{46}+\frac{250}{186}+\frac{-2121}{2323}+\frac{-125125}{143143}\)

\(=\frac{21}{23}+\frac{125}{93}+\frac{-21}{23}+\frac{-125}{143}\)

\(=\left(\frac{21}{23}+\frac{-21}{23}\right)+\frac{125}{93}+\frac{-125}{143}\)

\(=0+\frac{125}{93}+\frac{-125}{143}\)

\(=\frac{17875}{13299}+\frac{-11625}{13299}\)

\(=\frac{6250}{13299}\)

(Hình như câu b hơi sai)

\(a)\frac{4}{20}+\frac{16}{42}+\frac{6}{15}+\frac{-3}{5}+\frac{2}{21}+\frac{10}{21}+\frac{3}{20}\)

\(=\frac{1}{5}+\frac{8}{21}+\frac{2}{5}+\frac{-3}{5}+\frac{2}{21}+\frac{-10}{21}+\frac{3}{20}\)

\(=\left(\frac{1}{5}+\frac{2}{5}+\frac{-3}{5}\right)+\left(\frac{8}{21}+\frac{2}{21}+\frac{-10}{21}\right)+\frac{3}{20}\)

\(=0+0+\frac{3}{20}\)

\(=\frac{3}{20}\)

\(b)\frac{42}{46}+\frac{250}{186}+\frac{-2121}{2323}+\frac{-125125}{143143}\)

\(=\frac{21}{23}+\frac{125}{93}+\frac{-21}{23}+\frac{-125}{143}\)

\(=\left(\frac{21}{23}+\frac{-21}{23}\right)+\frac{125}{93}+\frac{-125}{143}\)

\(=0+\frac{125}{93}+\frac{-125}{143}\)

\(=\frac{125}{93}+\frac{-125}{143}\)

\(=\frac{6250}{13299}\)

Sao câu b) kq lại lớn vậy nhỉ , đề bài sai ? , hay mình làm sai ?

a, 4/20+16/42+6/15+-3/5+2/2+-10/21=1/5+8/21+2/5+-3/5+2/2+-16/21=(1/5+2/5+-3/5)+(8/21=-16/21)+1=-2/21+1=-19/21                       b,42/46+250/186+-2121/2323+-125125/143143                                                                                                                                =21/23+125/93+-21/23+-125/143                                                                                                                                                   =(21/23+-21/23)+(-125/143+125/93)                                                                                                                                              =6250/13299                                                             

a ) \(\frac{4}{20}+\frac{16}{42}+\frac{6}{15}+\frac{-3}{5}+\frac{2}{21}+\frac{-10}{21}+\frac{3}{20}\)

\(=\frac{4}{20}+\frac{8}{21}+\frac{2}{5}-\frac{3}{5}+\frac{2}{21}+\frac{-10}{21}+\frac{3}{20}\)

\(=\left(\frac{4}{20}+\frac{3}{20}\right)+\left(\frac{8}{21}+\frac{2}{21}-\frac{10}{21}\right)+\left(\frac{2}{5}-\frac{3}{5}\right)\)

\(=\frac{7}{20}+0+\frac{-1}{5}=\frac{7-4}{20}=\frac{3}{20}\)

b ) \(\frac{42}{46}+\frac{250}{186}+\frac{-2121}{2323}+\frac{-125125}{143143}\)

\(=\frac{21}{23}+\frac{-21}{23}+\frac{-125}{143}\)

\(=0+\frac{-125}{143}=-\frac{125}{143}\)

bài 2

a \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2003.2004}\)

=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2003}-\frac{1}{2004}\)
=\(1-\frac{1}{2004}=\frac{2003}{2004}\)

a ) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{2003.2004}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{2003}-\frac{1}{2004}\)

\(=1-\frac{1}{2004}=\frac{2003}{2004}\)

b ) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{2003.2005}\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{2003}-\frac{1}{2005}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{2005}\right)=\frac{1}{2}.\frac{2004}{2005}=\frac{1002}{2005}\)