a. 1/3; 1/6; 5/2; 3/2 bạn nha

c

a. Không gian mẫu gồm 36 phần tử:

Ω = {(i, j) | i, j = 1, 2, 3, 4, 5, 6 }

Trong đó (i, j) là kết quả "lần đầu xuất hiện mặt i chấm, lần sau xuất hiện mặt j chấm".

b. Phát biểu các biến cố dưới dạng mệnh đề:

A = {(6,1), (6,2), (6,3), (6,4), (6, 5), (6, 6)}

- Đây là biến cố "lần đầu xuất hiện mặt 6 chấm khi gieo con súc sắc".

B = {(2, 6), (6, 2), (3, 5), (5, 3), (4, 4)}

- Đây là biến cố " cả hai lần gieo có tổng số chấm bằng 8".

C = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}

- Đây là biến cố " kết quả của hai lần gieo là như nhau".

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Kho..................wa.....................troi.....................thi......................lanh.................ret.......................ai........................tich..........................ung.....................ho........................minh.....................cho....................do....................lanh

a. Ta có: a > b

4a > 4b ( nhân cả 2 vế cho 4)

4a - 3 > 4b - 3 (cộng cả 2 vế cho -3)

b. Ta có: a > b

-2a < -2b ( nhân cả 2 vế cho -2)

1 - 2a < 1 - 2b (cộng cả 2 vế cho 1)

d. Ta có: a < b 

-2a > -2b ( nhân cả 2 vế cho -2)

5 - 2a > 5 - 2b (cộng cả 2 vế cho 5)

 3/10 x x=2/5 

x = 2/5:3/10

x = 4/3

1/8 : x  = 1/2 

x = 1/8:1/2

x= 1/4

5/6 : 3/4  = 5/6x4/2= 10/9

1: 2/3= 1/1 x 3/2= 3/2

5/6 + 3/4  = 19/12

1 + 2/3= 1/1+2/3= 5/3

 5/6 - 3/4 = 1/12

1 - 2/3= 1/1-2/3= 1/3

5/6 x 3/4= 5/8

1/1x2/3= 2/3

A \ B = {0,1}

B \ A = {5;6}

(A\B) U (B\A) = {0;1;5;6}

=> A

a: \(\dfrac{6}{7}:\left(\dfrac{2}{5}\cdot\dfrac{6}{7}\right)\)

\(=\dfrac{6}{7}:\dfrac{12}{35}\)

\(=\dfrac{6}{7}\cdot\dfrac{35}{12}=\dfrac{6}{12}\cdot\dfrac{35}{7}=\dfrac{5}{2}\)

b: \(\dfrac{6}{7}+\dfrac{5}{7}:5-\dfrac{8}{9}\)

\(=\dfrac{6}{7}+\dfrac{1}{7}-\dfrac{8}{9}\)

\(=1-\dfrac{8}{9}=\dfrac{1}{9}\)

c: \(\dfrac{6}{7}+\dfrac{5}{8}\cdot\dfrac{1}{5}-\dfrac{3}{16}\cdot4\)

\(=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{4}\)

\(=\dfrac{48+7-42}{56}=\dfrac{13}{56}\)

d: \(\dfrac{-1}{6}+\dfrac{2}{3}\cdot\dfrac{-3}{4}+\dfrac{4}{5}\)

\(=-\dfrac{1}{6}-\dfrac{1}{2}+\dfrac{4}{5}\)

\(=\dfrac{-5-15+24}{30}=\dfrac{4}{30}=\dfrac{2}{15}\)

a) \(12\dfrac{1}{3}-\left(3\dfrac{3}{4}+4\dfrac{3}{4}\right)=\dfrac{37}{3}-\left(\dfrac{15}{4}+\dfrac{19}{4}\right)\)

\(=\dfrac{37}{3}-\dfrac{34}{4}=\dfrac{37}{3}-\dfrac{17}{2}=\dfrac{74}{6}-\dfrac{51}{6}=\dfrac{23}{6}\)

b) \(3\dfrac{5}{6}+2\dfrac{1}{6}.6=\dfrac{23}{6}+\dfrac{13}{6}.6=\dfrac{23}{6}+\dfrac{78}{6}=\dfrac{101}{6}\)

c) \(3\dfrac{1}{2}+4\dfrac{5}{7}-5\dfrac{5}{14}=\dfrac{7}{2}+\dfrac{33}{7}-\dfrac{75}{14}=\dfrac{49}{14}+\dfrac{66}{14}-\dfrac{75}{14}=-\dfrac{92}{14}=-\dfrac{46}{7}\)

d) \(4\dfrac{1}{2}+\dfrac{1}{2}:5\dfrac{1}{2}=\dfrac{9}{2}+\dfrac{1}{2}:\dfrac{11}{2}=\dfrac{9}{2}+\dfrac{1}{2}.\dfrac{2}{11}=\dfrac{9}{2}+\dfrac{1}{11}=\dfrac{99}{22}+\dfrac{2}{22}=\dfrac{101}{22}\)

a. \(12\dfrac{1}{3}-\left(3\dfrac{3}{4}+4\dfrac{3}{4}\right)=\dfrac{37}{3}-\left(\dfrac{15}{4}+\dfrac{19}{4}\right)\)

\(=\dfrac{37}{3}-\dfrac{34}{4}=\dfrac{23}{6}\)

\(b.3\dfrac{5}{6}+2\dfrac{1}{6}.6=\dfrac{23}{6}+13=\dfrac{101}{6}\)

\(c.3\dfrac{1}{2}+4\dfrac{5}{7}-5\dfrac{5}{14}=\dfrac{7}{2}+\dfrac{33}{7}-\dfrac{75}{14}=\dfrac{20}{7}\)

d  \(4\dfrac{1}{2}+\dfrac{1}{2}:5\dfrac{1}{2}\)

\(=\dfrac{9}{2}+\dfrac{1}{2}:\dfrac{11}{2}\)

\(=\dfrac{9}{2}+\dfrac{1}{11}\)

\(=\dfrac{101}{22}\)

101/22

a: \(=\dfrac{5}{9}-\dfrac{12+5}{15}:\dfrac{6}{5}\)

\(=\dfrac{5}{9}-\dfrac{17}{15}\cdot\dfrac{5}{6}=\dfrac{5}{9}-\dfrac{17}{18}=-\dfrac{7}{18}\)

b: \(=\left[\dfrac{6}{5}\left(3+\dfrac{3}{8}-6-\dfrac{5}{8}\right)\right]\cdot\dfrac{5}{6}=-3-\dfrac{1}{4}=-\dfrac{13}{4}\)

`a, 1/5 +5/9 + 4/5 +1/9 +3/9`

`=(1/5 +4/5)+(5/9 + 1/9+3/9)`

`= 5/5 + 9/9`

`=1+1`

`=2`

`b, 1/3 +1/6 +4/3 +2/6 + 2/3 + 5/6`

`=(1/3 + 4/3+2/3)+(1/6 +2/6+5/6)`

`= 7/3 + 8/6`

`=7/3 + 4/3`

`= 11/3`

`@ yL`