D

D

\(4\dfrac{3}{8}+5\dfrac{2}{3}=\dfrac{4\times8+3}{8}+\dfrac{5\times3+2}{3}\\ =\dfrac{35}{8}+\dfrac{17}{3}=\dfrac{35\times3+17\times8}{24}=\dfrac{241}{24}\\ ----\\ 2\dfrac{3}{8}+1\dfrac{1}{4}=\dfrac{2\times8+3}{8}+\dfrac{1\times4+1}{4}=\dfrac{19}{8}+\dfrac{5}{4}\\ =\dfrac{19+5\times2}{8}=\dfrac{29}{8}\\ ----\\ 2\dfrac{3}{8}-1\dfrac{1}{4}+5\dfrac{1}{3}=\left(2-1+5\right)+\left(\dfrac{3}{8}-\dfrac{1}{4}+\dfrac{1}{3}\right)\\ =6+\dfrac{3\times3-1\times6+1\times8}{24}\\ =6+\dfrac{11}{24}=6\dfrac{11}{24}\)

10

a, 35/8 + 17/3

= 241/24

b, 19/8 + 5/4 + 27/7

= 29/8 + 27/7

= 419/56

c, 19/8 - 5/4 + 27/7

= 9/8 + 27/7

= 279/56.

a) \(4\dfrac{3}{8}+5\dfrac{2}{3}\)

\(=\dfrac{35}{8}+\dfrac{17}{3}\)

\(=\dfrac{105}{24}+\dfrac{136}{24}\)

\(=\dfrac{241}{24}\)

b) \(2\dfrac{3}{8}+1\dfrac{1}{4}+3\dfrac{6}{7}\)

\(=\dfrac{19}{8}+\dfrac{5}{4}+\dfrac{27}{7}\)

\(=\dfrac{29}{8}+\dfrac{27}{7}\)

\(=\dfrac{419}{56}\)

c) \(2\dfrac{3}{8}-1\dfrac{1}{4}+5\dfrac{1}{3}\)

\(=\dfrac{19}{8}-\dfrac{5}{4}+\dfrac{16}{3}\)

\(=\dfrac{9}{8}+\dfrac{16}{3}\)

\(=\dfrac{155}{24}\)

d) \(\left(\dfrac{5}{2}+\dfrac{1}{3}\right):\left(1-\dfrac{1}{2}\right)\)

\(=\dfrac{17}{6}:\dfrac{1}{2}\)

\(=\dfrac{17}{6}\cdot2\)

\(=\dfrac{17}{3}\)

e) \(\left(\dfrac{5}{2}-\dfrac{1}{3}\right)\cdot\dfrac{9}{2}-\dfrac{6}{7}\)

\(=\dfrac{13}{6}\cdot\dfrac{9}{2}-\dfrac{6}{7}\)

\(=\dfrac{39}{4}-\dfrac{6}{7}\)

\(=\dfrac{249}{28}\)

a: =4+3/8+5+2/3

=9+9/24+16/24

=9+25/24

=216/24+25/24=241/24

b: \(=\dfrac{19}{8}+\dfrac{5}{4}+\dfrac{27}{7}=\dfrac{19+10}{8}+\dfrac{27}{7}\)

=27/7+29/8

=419/56

c: =2+3/8-1-1/4+5+1/3

=6+3/8-1/4+1/3

=6+3/8+1/12

=144/24+9/24+2/24

=155/24

d: =(15/6+2/6):1/2

=17/6*2

=17/3

e: =(15/6-2/6)*9/2-6/7

=13/6*9/2-6/7

=117/12-6/7

=249/28

đề sai

e. 1/3; 3/8; 1/2; 3/5

Chọn C

C

C

D

D

a) Ta có: \(A=\dfrac{16^8-1}{\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)

\(=\dfrac{2^{32}-1}{\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)

\(=\dfrac{2^{32}-1}{\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)

\(=\dfrac{2^{32}-1}{\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)

\(=\dfrac{2^{32}-1}{\left(2^{16}-1\right)\left(2^{16}+1\right)}\)

\(=\dfrac{2^{32}-1}{2^{32}-1}=1\)

b) Ta có: \(B=\dfrac{\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{9^{16}-1}\)

\(=\dfrac{\left(3^2-1\right)\cdot\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\cdot\left(3^{32}-1\right)}\)

\(=\dfrac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\cdot\left(3^{32}-1\right)}\)

\(=\dfrac{\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\left(3^{32}-1\right)}\)

\(=\dfrac{\left(3^{16}-1\right)\left(3^{16}+1\right)}{2\left(3^{32}-1\right)}=\dfrac{1}{2}\)

a) \(x-\dfrac{3}{4}=6\times\dfrac{3}{8}\)

\(x-\dfrac{3}{4}=\dfrac{9}{4}\)

=> \(x=\dfrac{9}{4}+\dfrac{3}{4}=3\)

b) \(\dfrac{7}{8}:x=3-\dfrac{1}{2}\)

\(\dfrac{7}{8}:x=\dfrac{5}{2}\)

=> \(x=\dfrac{7}{8}:\dfrac{5}{2}=\dfrac{7}{20}\)

c) \(x+\dfrac{1}{2}\times\dfrac{1}{3}=\dfrac{3}{4}\)

\(x+\dfrac{1}{6}=\dfrac{3}{4}\)

=> \(x=\dfrac{3}{4}-\dfrac{1}{6}=\dfrac{7}{12}\)

d) \(\dfrac{3}{2}\times\dfrac{4}{5}-x=\dfrac{2}{3}\)

\(\dfrac{6}{5}-x=\dfrac{2}{3}\)

=> \(x=\dfrac{6}{5}-\dfrac{2}{3}=\dfrac{8}{15}\)

e) \(x\times3\dfrac{1}{3}=3\dfrac{1}{3}:4\dfrac{1}{4}\)(?)

\(x\times\dfrac{10}{3}=\dfrac{40}{51}\)

=> \(x=\dfrac{40}{51}:\dfrac{10}{3}=\dfrac{4}{17}\)

f) \(5\dfrac{2}{3}:x=3\dfrac{2}{3}-2\)

\(\dfrac{17}{3}:x=\dfrac{5}{3}\)

=> \(x=\dfrac{17}{3}:\dfrac{5}{3}=\dfrac{17}{5}\)

a: =>x-3/4=18/8=9/4

=>x=9/4+3/4=12/4=3

b: =>7/8:x=5/2

=>x=7/8:5/2=7/8*2/5=14/40=7/20

c: x+1/2*1/3=3/4

=>x+1/6=3/4

=>x=3/4-1/6=9/12-2/12=7/12

d: =>12/10-x=2/3

=>6/5-x=2/3

=>x=6/5-2/3=18/15-10/15=8/15

e: =>x*10/3=10/3:17/4=10/3*4/17

=>x=4/17

f: =>17/3:x=13/3-5/2=26/6-15/6=11/6

=>x=17/3:11/6=17/3*6/11=34/11

Ta có:

Tập hợp A:
\(A=\left\{1;2;3;5;8\right\}\)

Tập hợp B:

\(B=\left\{-1;0;1;5;9\right\}\)

Mà: \(A\cup B\)

\(\Rightarrow A\cup B=\left\{-1;0;1;2;3;5;8;9\right\}\)

⇒ Chọn B

Chọn B