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

a, -51   b, -40

Haha

a, \(\dfrac{7}{22}\) - \(\dfrac{15}{23}\) + \(\dfrac{2022}{2023}\) - \(\dfrac{8}{23}\) + \(\dfrac{15}{22}\)

= ( \(\dfrac{7}{22}\) + \(\dfrac{15}{22}\)) - ( \(\dfrac{15}{23}+\dfrac{18}{23}\)) + \(\dfrac{2022}{2023}\)

= \(\dfrac{22}{22}\) - \(\dfrac{23}{23}\) + \(\dfrac{2022}{2023}\)

= 1 - 1 + \(\dfrac{2022}{2023}\)

= \(\dfrac{2022}{2023}\) 

b, - \(\dfrac{2}{11}\) + 5\(\dfrac{5}{6}\) ( 14\(\dfrac{1}{5}\) - 11\(\dfrac{1}{5}\)): 5\(\dfrac{1}{2}\)

= - \(\dfrac{2}{11}\) + \(\dfrac{35}{6}\) ( \(\dfrac{71}{5}\) - \(\dfrac{56}{5}\)) : \(\dfrac{11}{2}\)

= - \(\dfrac{2}{11}\) + \(\dfrac{35}{6}\) . \(\dfrac{15}{5}\) : \(\dfrac{11}{2}\)

= - \(\dfrac{2}{11}\) + \(\dfrac{35}{2}\) \(\times\) \(\dfrac{2}{11}\)

= - \(\dfrac{2}{11}\) + \(\dfrac{35}{11}\)

= \(\dfrac{33}{11}\)

= 3 

c, 2000 + { 20 - [ 4.2022⁰ - (3² + 5):2] }

= 2000 + { 20 - [ 4.1 - (9+5):2]}

= 2000 + { 20 - [ 4 - 14 : 2 ]}

= 2000 + { 20 - [ 4 -7]}

= 2000 + { 20 - (-3)}

= 2000 + 23

= 2023

a: \(\frac{7}{22}+\frac{-15}{23}+\frac{2022}{2023}+\frac{-8}{23}+\frac{15}{22}\)

\(=\left(\frac{7}{22}+\frac{15}{22}\right)+\left(-\frac{15}{23}-\frac{8}{23}\right)+\frac{2022}{2023}\)

\(=1-1+\frac{2022}{2023}=\frac{2022}{2023}\)

b: \(-\frac{2}{11}+5\frac56\left(14\frac15-11\frac15\right):5\frac12\)

\(=-\frac{2}{11}+\frac{35}{6}\left(14+\frac15-11-\frac15\right):\frac{11}{2}\)

\(=\frac{-2}{11}+\frac{35}{6}\cdot\frac{2}{11}\cdot3=\frac{-2}{11}+\frac{35}{11}=\frac{33}{11}=3\)

c: \(2000+\left\lbrace20-\left\lbrack4\cdot2022^0-\left(3^2+5\right):2\right\rbrack\right\rbrace\)

\(=2000+20-\left\lbrack4\cdot1-14:2\right\rbrack\)

=2020-(4-7)

=2020-(-3)

=2023

a: \(17\cdot85+15\cdot17-120\)

\(=17\left(85+15\right)-120\)

=1700-120

=1580

b: \(2^3\cdot17-2^3\cdot14\)

\(=2^3\left(17-14\right)\)

\(=8\cdot3=24\)

c: \(20-\left\lbrack30-\frac{\left(5-1\right)^2}{2}\right\rbrack\)

\(=20-30+\frac{4^2}{2}\)

\(=-10+\frac{16}{2}=-10+8=-2\)

d: \(80-\left(4\cdot5^2-3\cdot2^2\right)\)

\(=80-\left(4\cdot25-3\cdot4\right)\)

=80-100+12

=92-100

=-8

e: \(27\cdot77+24\cdot27-27\)

\(=27\left(77+24-1\right)\)

\(=27\cdot100=2700\)

f: \(174:\left\lbrace2\cdot\left\lbrack36+\left(4^2-23\right)\right\rbrack\right\rbrace\)

\(=\frac{174}{2\cdot\left(36+16-23\right)}=\frac{87}{52-23}=\frac{87}{29}=3\)

a: \(\left(\frac45-\frac{13}{20}\right):\frac35\)

\(=\left(\frac{16}{20}-\frac{13}{20}\right)\cdot\frac53\)

\(=\frac{3}{20}\cdot\frac53=\frac{5}{20}=\frac14\)

b: \(197-13\cdot8+768\)

=197-104+768

=93+768

=861

c: \(20,7+1,47:7-23\%\cdot5\)

=20,7+0,21-1,15

=19,76

d: \(\frac{21}{77}\cdot\frac12+\frac{4}{11}\cdot\frac16+\frac{5}{11}\cdot\frac13\)

\(=\frac{3}{11}\cdot\frac12+\frac{2}{11}\cdot\frac13+\frac{5}{11}\cdot\frac13=\frac{3}{11}\cdot\frac12+\frac{7}{11}\cdot\frac13=\frac{3}{22}+\frac{7}{33}\)

\(=\frac{9}{66}+\frac{14}{66}=\frac{23}{66}\)

Vì  ( − 1 ) . ( − 2 ) . ( − 3 ) ... ( − 20 ) > 0 ( − 3 ) . ( − 4 ) . ( − 5 ) ... ( − 23 ) < 0 ⇒ ( − 1 ) . ( − 2 ) . ( − 3 ) ... ( − 20 ) > ( − 3 ) . ( − 4 ) . ( − 5 ) ... ( − 23 )

Ủa bạn ??

Zậy là tính cả gì vậy

Bài 1:

a: \(\frac{13}{20}<1\)

\(\frac{19}{20}<1\)

\(\frac{17}{23}<1\)

Do đó: \(\frac{13}{20}+\frac{19}{20}+\frac{17}{23}<1+1+1=3\)

hay \(\frac{13}{20}+\frac{19}{20}+\frac{17}{23}<3\)

b: \(\frac75>1;\frac54>1;\frac98>1;\frac32>1\)

Do đó; \(\frac75+\frac54+\frac98+\frac32>1+1+1+1\)

=>\(\frac75+\frac54+\frac98+\frac32>4\)

Bài 2:

\(\frac87+\frac{11}{13}+\frac{23}{30}+\frac{39}{45}<4\)