1A

2B

3B

4D

5A

...dài quá, em tách ra dùm a =')

Câu 1: D

Câu 2: C

A

a

A

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

`#3107.101107`

A.

`12 + (-2).8`

`= 12 + (-16)`

`= 12 - 16`

`= -4`

Vì `-4 < 23 =>` đáp án A không thỏa mãn dk

B.

`8 - 4 + 37`

`= 4 + 37`

`= 41`

Vì `41 > 23 =>` đáp án B không thỏa mãn dk

C.

`7.4 + (-3)`

`= 28 + (-3)`

`= 28 - 3`

`= 25`

Vì `25 > 23 =>` đáp án C không thỏa mãn

D.

`9.8 - 72`

`= 72 - 72`

`= 0`

Vì `0 < 23 =>` đáp án D không thỏa mãn.

Vậy, không có đáp án nào thỏa mãn dk.

C7: A

C8:A

C9:D

A. -1;-3;-89;-98

A. 6

D. 6

~hok tốt

Sao cx gặp bài phân số không -_-

\(a)\frac{19}{21}-\frac{1}{3}+\frac{3}{7}=\frac{19}{21}-(\frac{1}{3}+\frac{3}{7})=\frac{19}{21}-\frac{16}{21}=\frac{3}{21}\)

\(b)1+\frac{2}{3}-\frac{5}{3}=\frac{3}{3}+\frac{2}{3}-\frac{5}{3}=\frac{5}{3}-\frac{5}{3}=0\)

\(c)\frac{7}{12}+\frac{1}{3}-\frac{3}{4}=\frac{7}{12}+\frac{4}{12}-\frac{9}{12}=\frac{7+4-9}{12}=\frac{2}{12}=\frac{1}{6}\)

\(d)\frac{5}{7}-\frac{4}{7}\cdot\frac{1}{4}=\frac{5}{7}-\frac{1}{7}=\frac{4}{7}\)

\(e)\frac{7}{12}\cdot\frac{4}{3}+\frac{2}{9}=\frac{7}{9}+\frac{2}{9}=\frac{9}{9}=1\)

A. \(\frac{19}{21}-\frac{1}{3}+\frac{3}{7}=\frac{4}{7}+\frac{3}{7}=\frac{7}{7}=1\)

B. \(1+\frac{2}{3}-\frac{5}{3}=\frac{5}{3}-\frac{5}{3}=0\)

C. \(\frac{7}{12}+\frac{1}{3}-\frac{3}{4}=\frac{11}{12}-\frac{3}{4}=\frac{1}{6}\)

D. \(\frac{5}{7}-\frac{4}{7}\times\frac{1}{4}=\frac{5}{7}-\frac{1}{7}=\frac{4}{7}\)

E. \(\frac{7}{12}\times\frac{4}{3}+\frac{2}{9}=\frac{7}{9}+\frac{2}{9}=\frac{9}{9}=1\)

                       \(Linh\)\(Yukiko\)                          

Sửa lại câu a : \(\frac{19}{21}-\frac{1}{3}+\frac{3}{7}=\frac{19}{21}-\frac{7}{21}+\frac{9}{21}=\frac{19-7+9}{21}=\frac{21}{21}=1\)

A

A

A

a: \(=\dfrac{7}{2}\left(-\dfrac{3}{4}+\dfrac{5}{13}-\dfrac{9}{4}-\dfrac{8}{13}\right)=\dfrac{7}{2}\cdot\left(-3-\dfrac{3}{13}\right)=\dfrac{7}{2}\cdot\dfrac{-42}{13}=\dfrac{-147}{13}\)

b: \(=-12+\dfrac{8}{9}-\dfrac{5}{18}=\dfrac{-216}{18}+\dfrac{16}{18}-\dfrac{5}{18}=\dfrac{-205}{18}\)

c: \(=\dfrac{45}{4}-\dfrac{19}{7}-\dfrac{21}{4}=6-\dfrac{19}{7}=\dfrac{23}{7}\)

d: \(=\dfrac{-1}{4}\left(\dfrac{152}{11}+\dfrac{68}{11}\right)=\dfrac{-1}{4}\cdot20=-5\)

2.D 

3.11/-6

\(\dfrac{3}{5}:\left(\dfrac{1}{4}.\dfrac{7}{5}\right)=\dfrac{3}{5}:\dfrac{7}{20}\)

\(=\dfrac{60}{35}=\dfrac{12}{7}\)

=> D

Câu 3 : \(\dfrac{-6}{11}=>\dfrac{-11}{6}\)

2.d
3.\(\dfrac{-6}{11}\)=\(\dfrac{6}{11}\)