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)

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

\(\begin{array}{l}\left[ {\left( { - 3} \right) + 4} \right] + 2 = \left( {4 - 3} \right) + 2\\ = 1 + 2 = 3\end{array}\)

\(\begin{array}{l}\left( { - 3} \right) + \left( {4 + 2} \right) = \left( { - 3} \right) + 6\\ = 6 - 3 = 3\end{array}\)

\(\begin{array}{l}\left[ {\left( { - 3} \right) + 2} \right] + 4 =  - \left( {3 - 2} \right) + 4\\ =  - 1 + 4 = 3\end{array}\)

a) \({\left[ {{{\left( { - 2} \right)}^2}} \right]^3} = {\left( { - 2} \right)^2}.{\left( { - 2} \right)^2}.{\left( { - 2} \right)^2} = {\left( { - 2} \right)^{2 + 2 + 2}} = {\left( { - 2} \right)^6}\)

Vậy \({\left[ {{{\left( { - 2} \right)}^2}} \right]^3}\) = \({\left( { - 2} \right)^6}\)        

b) \({\left[ {{{\left( {\frac{1}{2}} \right)}^2}} \right]^2} = {\left( {\frac{1}{2}} \right)^2}.{\left( {\frac{1}{2}} \right)^2} = {\left( {\frac{1}{2}} \right)^4}\)

Vậy \({\left[ {{{\left( {\frac{1}{2}} \right)}^2}} \right]^2}\) = \({\left( {\frac{1}{2}} \right)^4}\).

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

a: Vì 0,2<1

nên hàm số \(y=\left(0,2\right)^x\) nghịch biến trên R

mà -3<-2

nên \(\left(0,2\right)^{-3}>\left(0,2\right)^{-2}\)

b: Vì \(0< \dfrac{1}{3}< 1\)

nên hàm số \(y=\left(\dfrac{1}{3}\right)^x\) nghịch biến trên R

mà \(2000< 2004\)

nên \(\left(\dfrac{1}{3}\right)^{2000}>\left(\dfrac{1}{3}\right)^{2004}\)

c: Vì 3,2>1

nên hàm số \(y=\left(3,2\right)^x\) đồng biến trên R

mà \(1,5< 1,6\)

nên \(\left(3,2\right)^{1,5}< \left(3,2\right)^{1,6}\)

d: Vì \(0< 0,5< 1\)

nên hàm số \(y=\left(0,5\right)^x\) nghịch biến trên R

mà -2021>-2023

nên \(\left(0,5\right)^{-2021}< \left(0,5\right)^{-2023}\)

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

\(a.\)

\(\dfrac{3}{10}:\left(-\dfrac{2}{3}\right)=\dfrac{3}{10}\cdot\dfrac{-3}{2}=-\dfrac{9}{20}\)

\(b.\)

\(\left(-\dfrac{7}{12}\right):\left(-\dfrac{5}{6}\right)=\left(-\dfrac{7}{12}\right)\cdot\left(-\dfrac{6}{5}\right)=\dfrac{\left(-7\right)\cdot\left(-6\right)}{12\cdot5}=\dfrac{7}{10}\)

\(c.\)

\(\left(-15\right):-\dfrac{9}{10}=\left(-15\right)\cdot-\dfrac{10}{9}=\dfrac{150}{9}=\dfrac{50}{3}\)

a) \(\dfrac{3}{10}:\dfrac{-2}{3}=\dfrac{3}{10}.\dfrac{-3}{2}=\dfrac{3.-3}{10.2}=\dfrac{-9}{20}\) 

b) \(\dfrac{-7}{12}:\dfrac{-5}{6}=\dfrac{-7}{12}.\dfrac{-6}{5}=\dfrac{-7.-6}{12.5}=\dfrac{7}{10}\) 

c)\(-15:\dfrac{-9}{10}=-15.\dfrac{-10}{9}=\dfrac{-15.-10}{9}=\dfrac{50}{3}\)

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

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