\(\left( { - 1} \right) + \left( { - 3} \right) =  - \left( {1 + 3} \right) =  - 4\)

\(\left( { - 3} \right) + \left( { - 1} \right) =  - \left( {3 + 1} \right) =  - 4\)

\( \Rightarrow \left( { - 1} \right) + \left( { - 3} \right) = \left( { - 3} \right) + \left( { - 1} \right)\)

\(\left( { - 7} \right) + \left( { + 6} \right) =  - \left( {7 - 6} \right) =  - 1\)

\(\left( { + 6} \right) + \left( { - 7} \right) =  - \left( {7 - 6} \right) =  - 1\)

\( \Rightarrow \left( { - 7} \right) + \left( { + 6} \right) = \left( { + 6} \right) + \left( { - 7} \right)\)

a) \( - \left( {4 + 7} \right) =  - 11\)

\(\begin{array}{l}\left( { - 4 - 7} \right) = \left( { - 4} \right) + \left( { - 7} \right)\\ =  - \left( {4 + 7} \right) =  - 11\\ \Rightarrow \left( { - 4 - 7} \right) =  - \left( {4 + 7} \right)\end{array}\)

b)

\(\begin{array}{l} - \left( {12 - 25} \right) =  - \left[ {12 + \left( { - 25} \right)} \right]\\ =  - \left[ { - \left( {25 - 12} \right)} \right] =  - \left( { - 13} \right) = 13\end{array}\)

\(\begin{array}{l}\left( { - 12 + 25} \right) = 25 - 12 = 13\\ \Rightarrow  - \left( {12 - 25} \right) = \left( { - 12 + 25} \right)\end{array}\)

c)

\(\begin{array}{l} - \left( { - 8 + 7} \right) =  - \left[ { - \left( {8 - 7} \right)} \right] =  - \left( { - 1} \right) = 1\\\left( {8 - 7} \right) = 1\\ \Rightarrow  - \left( { - 8 + 7} \right) = \left( {8 - 7} \right)\end{array}\)

d)

\(\begin{array}{l} + \left( { - 15 - 4} \right) =  + \left[ {\left( { - 15} \right) + \left( { - 4} \right)} \right]\\ =  + \left[ { - \left( {15 + 4} \right)} \right] =  + \left( { - 19} \right) =  - 19\\\left( { - 15 - 4} \right) = \left( { - 15} \right) + \left( { - 4} \right)\\ =  - \left( {15 + 4} \right) =  - 19\\ \Rightarrow  + \left( { - 15 - 4} \right) = \left( { - 15 - 4} \right)\end{array}\)

e)

\(\begin{array}{l} + \left( {23 - 12} \right) =  + 11 = 11\\\left( {23 - 12} \right) = 11\\ \Rightarrow  + \left( {23 - 12} \right) = \left( {23 - 12} \right)\end{array}\)

a: \(log_2\left(mn\right)=log_2\left(2^7\cdot2^3\right)=7+3=10\)

 \(log_2m+log_2n=log_22^7+log_22^3=7+3=10\)

=>\(log_2\left(mn\right)=log_2m+log_2n\)

b: \(log_2\left(\dfrac{m}{n}\right)=log_2\left(\dfrac{2^7}{2^3}\right)=7-3=4\)

\(log_2m-log_2n=log_22^7-log_22^3=7-3=4\)

=>\(log_2\left(\dfrac{m}{n}\right)=log_2m-log_2n\)

a) \(\log_2\left(mn\right)=\log_2\left(2^7.2^3\right)=\log_22^{7+3}=\log_22^{10}=10.\log_22=10.1=10\)

\(\log_2m+\log_2n=\log_22^7+\log_22^3=7\log_22+3\log_22=7.1+3.1=7+3=10\)

b) \(\log_2\left(\dfrac{m}{n}\right)=\log_2\dfrac{2^7}{2^3}=\log_22^4=4.\log_22=4.1=4\)

\(\log_2m-\log_2n=\log_22^7-\log_22^3=7.\log_22-3\log_22=7.1-3.1=4\)

a) \(\frac{3}{4} + \left( {\frac{1}{2} - \frac{1}{3}} \right) = \frac{9}{{12}} + \left( {\frac{6}{{12}} - \frac{4}{{12}}} \right) = \frac{9}{{12}} + \frac{2}{{12}} = \frac{{11}}{{12}}\)

\(\frac{3}{4} + \frac{1}{2} - \frac{1}{3} = \frac{9}{{12}} + \frac{6}{{12}} - \frac{4}{{12}} = \frac{{15}}{{12}} - \frac{4}{{12}} = \frac{{11}}{{12}}\)

Vậy \(\frac{3}{4} + \left( {\frac{1}{2} - \frac{1}{3}} \right)\) = \(\frac{3}{4} + \frac{1}{2} - \frac{1}{3}\)    

b)\(\frac{2}{3} - \left( {\frac{1}{2} + \frac{1}{3}} \right) = \frac{4}{6} - \left( {\frac{3}{6} + \frac{2}{6}} \right) = \frac{4}{6} - \frac{5}{6} = \frac{{ - 1}}{6}\)

 \(\frac{2}{3} - \frac{1}{2} - \frac{1}{3} = \frac{4}{6} - \frac{3}{6} - \frac{2}{6} = \frac{1}{6} - \frac{2}{6} = \frac{{ - 1}}{6}\)

Vậy \(\frac{2}{3} - \left( {\frac{1}{2} + \frac{1}{3}} \right)\)=\(\frac{2}{3} - \frac{1}{2} - \frac{1}{3}\).

`#3107`

`a)`

`3/4 + (1/2 - 1/3)`

`= 3/4 + (3/6 - 2/6)`

`= 3/4 + 1/6`

`= 11/12`

`3/4 + 1/2 - 1/3`

`= 9/12 + 6/12 - 4/12`

`= (9 + 6 - 4)/12`

`= 11/12`

Vì `11/12 = 11/12`

`=> 3/4 + (1/2 - 1/3) = 3/4 + 1/2 - 1/3`

`b)`

`2/3 - (1/2 + 1/3)`

`= 2/3 - (3/6 + 2/6)`

`= 2/3 - 5/6`

`= -1/6`

`2/3 - 1/2 - 1/3`

`= 4/6 - 3/6 - 2/6`

`= (4 - 3 - 2)/6`

`= -1/6`

Vì `-1/6 = -1/6`

`=> 2/3 - (1/2 + 1/3) = 2/3 - 1/2 - 1/3`

a)\(\left( { - 35,1} \right).\left( { - 64} \right):13 \approx \left( { - 35} \right).\left( { - 64} \right):13 \approx 172\)

b)\(\left( { - 8,8} \right).\left( { - 4,1} \right):{\rm{ }}2,6 \approx ( - 9).( - 4):3 = 12\)

c) \(7,9.\left( { - 73} \right):\left( { - 23} \right) \approx 8.( - 73):( - 23) \approx 25\).

toàn hỏi lung tung. lớp 6 mà còn ko biết làm mấy bài toán vớ vẩn kia

Số kết quả thuận lợi cho biến cố \(A \cup B\) là \(5 + 12 = 17\).

\(P\left( A \right) = \frac{{n\left( A \right)}}{{n\left( \Omega \right)}} = \frac{5}{{n\left( \Omega \right)}};P\left( B \right) = \frac{{n\left( B \right)}}{{n\left( \Omega\right)}} = \frac{{12}}{{n\left( \Omega\right)}};P\left( {A \cup B} \right) = \frac{{n\left( {A \cup B} \right)}}{{n\left( \Omega\right)}} = \frac{{17}}{{n\left( \Omega\right)}}\)

\( \Rightarrow P\left( A \right) + P\left( B \right) = P\left( {A \cup B} \right)\)

a) \(37+\left(-27\right)=10=\left(-27\right)+37\)

b) \(16+(-16)=0=(-105)+105\)

a)37+(-27)=10và -27+37=10

vậy 37+(-27)=-27+37

b)16+(-16)=0 và -105+105=0

vậy 2 kết quả này bằng nhau

a) 37 + (-27) = 37 – 27 = 10

(-27) + 37 = 37 – 27 = 10

Nhận xét: Nếu đổi chỗ các số hạng thì tổng không thay đổi.

b) 16 + (-16) = 16 – 16 = 0

(-105) + 105 = 105 – 105 = 0

Nhận xét: Tổng hai số nguyên đối nhau luôn bằng 0.

\(B=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

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

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

                \(.........\)

\(=\frac{1}{2}\left(3^{32}-1\right)\)\(< \)\(3^{32}-1\)\(=\)\(A\)

Vậy  \(B< A\)

 A=1.853020189*10 \(^{15}\)

B= 9.265100944*10\(^{15}\)

tự so sánh

Xét B ta có:

\(2B=2\left(3+1\right).\left(3^2+1\right).\left(3^4+1\right).\left(3^8+1\right).\left(3^{16}+1\right)\)

\(2B=\left(3-1\right)\left(3+1\right).\left(3^2+1\right).\left(3^4+1\right).\left(3^8+1\right).\left(3^{16}+1\right)\)

\(2B=\left(3^2-1\right).\left(3^2+1\right).\left(3^4+1\right).\left(3^8+1\right).\left(3^{16}+1\right)\)

\(2B=\left(3^4-1\right).\left(3^4+1\right).\left(3^8+1\right).\left(3^{16}+1\right)\)

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

\(2B=\left(3^{16}-1\right).\left(3^{16}+1\right)\)

\(2B=3^{32}-1\)

\(B=\frac{3^{32}-1}{2}< A=3^{32}-1\)

Vậy B < A