1+1=3 :)))

a)

\(\begin{array}{l}{\left( {\frac{{ - 1}}{2}} \right)^5} = \frac{{{{\left( { - 1} \right)}^5}}}{{{2^5}}} = \frac{{ - 1}}{{32}};\\{\left( {\frac{{ - 2}}{3}} \right)^4} = \frac{{{{\left( { - 2} \right)}^4}}}{{{3^4}}} = \frac{{16}}{{81}};\\{\left( { - 2\frac{1}{4}} \right)^3} = {\left( {\frac{{ - 9}}{4}} \right)^3} = \frac{{{{\left( { - 9} \right)}^3}}}{{{4^3}}} = \frac{{-729}}{{64}};\\{\left( { - 0,3} \right)^5} = {\left( {\frac{{ - 3}}{{10}}} \right)^5} = \frac{{ - 243}}{{100000}};\\{\left( { - 25,7} \right)^0} = 1\end{array}\)

b)

\(\begin{array}{l}{\left( { - \frac{1}{3}} \right)^2} = \frac{1}{9};\\{\left( { - \frac{1}{3}} \right)^3} = \frac{{ - 1}}{{27}};\\{\left( { - \frac{1}{3}} \right)^4} = \frac{1}{{81}};\\{\left( { - \frac{1}{3}} \right)^5} = \frac{{ - 1}}{{243}}.\end{array}\)

Nhận xét:

+ Luỹ thừa của một số hữu tỉ âm với số mũ chẵn là một số hữu tỉ dương.

+  Luỹ thừa của một số hữu tỉ âm với số mũ lẻ là một số hữu tỉ âm.

Đặt A = 1/2 - 1/3 - 2/3 + 1/4 + 2/4 + 3/4 - 1/5 - 2/5 - 3/5 - 4/5 + ... + 1/10 + ...+ 9/10

A = 1/2 - ( 1/3 + 2/3) + (1/4 + 2/4 + 3/4) - ( 1/5 + 2/5 + 3/5 + 4/5) + ( 1/6 + 2/6 + ...  + 5/6) - ( 1/7 + 2/7 + ... + 6/7) + ( 1/8 + 2/8 + ... + 7/8) - ( 1/9 + 2/9 + ... + 8/9)

A = 1/2 - 1 + [( 1/4 + 3/4) + 2/4] - [(1/5 + 4/5) + (2/5 + 3/5)] + [(1/6+5/6) + ( 2/6 + 4/6) + 3/6] - [(1/7 + 6/7) + (2/7 + 5/7) + (3/7 + 4/7)] + [(1/8 + 7/8) + (2/8 + 6/8) + (3/8 + 5/8) + 4/8)] - [(1/9 + 8/9) + (2/9 + 7/9) + (3/9 + 6/9) + (4/9 + 5/9)] + [(1/10 + 9/10) + ( 2/10 + 8/10) + ( 3/10 + 7/10) + ( 4/10 + 6/10) + 5/10]

A = 1/2 - 1 + ( 1 + 1/2) - 2 + ( 2 + 1/2) - 3 + ( 3 + 1/2) - 4 + ( 4 + 1/2) 

A = 1/2 + 1/2 + 1/2 + 1/2 + 1/2 

A = 1/2 × 5 = 5/2

Đặt A = 1/2 - 1/3 - 2/3 + 1/4 + 2/4 + 3/4 - 1/5 - 2/5 - 3/5 - 4/5 + ... + 1/10 + ...+ 9/10

A = 1/2 - ( 1/3 + 2/3) + (1/4 + 2/4 + 3/4) - ( 1/5 + 2/5 + 3/5 + 4/5) + ( 1/6 + 2/6 + ...  + 5/6) - ( 1/7 + 2/7 + ... + 6/7) + ( 1/8 + 2/8 + ... + 7/8) - ( 1/9 + 2/9 + ... + 8/9)

A = 1/2 - 1 + [( 1/4 + 3/4) + 2/4] - [(1/5 + 4/5) + (2/5 + 3/5)] + [(1/6+5/6) + ( 2/6 + 4/6) + 3/6] - [(1/7 + 6/7) + (2/7 + 5/7) + (3/7 + 4/7)] + [(1/8 + 7/8) + (2/8 + 6/8) + (3/8 + 5/8) + 4/8)] - [(1/9 + 8/9) + (2/9 + 7/9) + (3/9 + 6/9) + (4/9 + 5/9)] + [(1/10 + 9/10) + ( 2/10 + 8/10) + ( 3/10 + 7/10) + ( 4/10 + 6/10) + 5/10]

A = 1/2 - 1 + ( 1 + 1/2) - 2 + ( 2 + 1/2) - 3 + ( 3 + 1/2) - 4 + ( 4 + 1/2) 

A = 1/2 + 1/2 + 1/2 + 1/2 + 1/2 

A = 1/2 × 5 = 5/2

(A) \(\frac{2}{3} + \frac{{ - 4}}{6} = \frac{4}{6} + \frac{{ - 4}}{6} = 0\) => A sai

(B) \(\frac{2}{3}.\frac{{ - 1}}{5} = \frac{{ - 2}}{{15}}\) mà \(\frac{{3 - 2}}{5} = \frac{1}{5}\) => B sai

(C) \(\frac{2}{3} - \frac{3}{5} = \frac{{10}}{{15}} - \frac{9}{{15}} = \frac{1}{{15}}\) => C đúng

(D) \(\frac{3}{5}:\frac{3}{{ - 5}} = \frac{3}{5}.\frac{{ - 5}}{3} = \frac{{ - 15}}{{15}} =  - 1\) => D sai

=> Chọn C.

\(\left(\frac{3}{5}-\frac{1}{2}+\frac{7}{3}\right)-\left(\frac{1}{3}-\frac{5}{2}+\frac{1}{5}\right)-\left(-\frac{3}{5}+3-\frac{5}{3}\right)\)

\(=\frac{3}{5}-\frac{1}{2}+\frac{7}{3}-\frac{1}{3}+\frac{5}{2}-\frac{1}{5}+\frac{3}{5}-3+\frac{5}{3}\)

\(=\frac{3}{5}+\frac{-1}{2}+\frac{7}{3}+\frac{-1}{3}+\frac{5}{2}+\frac{-1}{5}+\frac{3}{5}+\left(-3\right)+\frac{5}{3}\)

\(=\left(\frac{3}{5}+\frac{-1}{5}+\frac{3}{5}\right)+\left(\frac{-1}{2}+\frac{5}{2}\right)+\left(\frac{7}{3}+\frac{-1}{3}+\frac{5}{3}\right)+\left(-3\right)\)

\(=\frac{5}{2}+\frac{4}{2}+\frac{11}{3}+\left(-3\right)\)

\(=1+2+\frac{11}{3}+\left(-3\right)\)

\(=0+\frac{11}{3}\)

\(=\frac{11}{3}\)

\(\left(\frac{3}{5}-\frac{1}{2}+\frac{7}{3}\right)-\left(\frac{1}{3}-\frac{5}{2}+\frac{1}{5}\right)-\left(-\frac{3}{5}+3-\frac{5}{3}\right)\)

\(=\frac{3}{5}-\frac{1}{2}+\frac{7}{3}-\frac{1}{3}+\frac{5}{2}-\frac{1}{5}+\frac{3}{5}-3+\frac{5}{3}\)

\(=\left(\frac{3}{5}-\frac{1}{5}+\frac{3}{5}\right)+\left(-\frac{1}{2}+\frac{5}{2}\right)+\left(\frac{7}{3}-\frac{1}{3}+\frac{5}{3}\right)-3\)

\(=1+2+\frac{11}{3}-3\)

\(=\left(1+2-3\right)+\frac{11}{3}\)

\(=0+\frac{11}{3}\)

=\(\frac{11}{3}\)

(Bài làm có gì ko hiểu bn cứ hỏi mk nhé  ^...^ ^_^)

Theo đề ta có:

\(\left(5+\frac{1}{5}-\frac{2}{9}\right)-\left(2-\frac{1}{23}-2\frac{3}{5}+\frac{5}{6}\right)\)\(-\left(8-\frac{2}{3}-\frac{1}{18}\right)\)

= \(5+\)\(\frac{1}{5}-\frac{2}{9}\)-\(2+\frac{1}{23}+2+\frac{3}{5}+\frac{5}{6}-8+\frac{2}{3}-\frac{1}{18}\)

=\(\left(5+2-8\right)+\left(\frac{1}{5}+\frac{3}{5}\right)-\left(\frac{2}{9}-\frac{5}{6}-\frac{2}{3}+\frac{1}{18}\right)+\frac{1}{23}\)

=  -1 +\(\frac{4}{5}\)\(-\frac{-11}{9}\)+\(\frac{1}{23}\)

= -1 +\(\frac{4}{5}+\frac{11}{9}+\frac{1}{23}\)

\(\left(5+\frac{1}{5}-\frac{2}{9}\right)-\left(2-\frac{1}{23}-2\frac{3}{5}+\frac{5}{6}\right)-\left(8-\frac{2}{3}-\frac{1}{18}\right)\)

= \(5+\frac{1}{5}-\frac{2}{9}-2+\frac{1}{23}+2+\frac{3}{5}-\frac{5}{6}-8+\frac{2}{3}+\frac{1}{18}\)

= \(\left(5-8\right)+\left(\frac{1}{5}+\frac{3}{5}\right)-\left(\frac{2}{9}-\frac{1}{18}-\frac{2}{3}\right)-\left(2-2\right)+\frac{1}{23}-\frac{5}{6}\)

= \(\left(-3\right)+\frac{4}{5}+\frac{1}{2}+\frac{1}{23}-\frac{5}{6}\)

= \(\left(\left(-3\right)+\frac{4}{5}+\frac{1}{2}-\frac{5}{6}\right)+\frac{1}{23}\)

= \(-\frac{38}{15}+\frac{1}{23}\)

= \(-\frac{859}{345}\)

a) \(\frac{{ - 3}}{7}.\frac{2}{5} + \frac{2}{5}.\left( { - \frac{5}{{14}}} \right) - \frac{{18}}{{35}}\)

\(\begin{array}{l} = \frac{2}{5}.\left( {\frac{{ - 3}}{7} + \frac{{ - 5}}{{14}}} \right) - \frac{{18}}{{35}}\\ = \frac{2}{5}.\left( {\frac{{ - 6}}{{14}} + \frac{{ - 5}}{{14}}} \right) - \frac{{18}}{{35}}\\ = \frac{2}{5}.\frac{{ - 11}}{{14}} - \frac{{18}}{{35}} = \frac{{ - 11}}{{35}} - \frac{{18}}{{35}} =  \frac{{ -29}}{{35}}\end{array}\)

b) \(\left( {\frac{2}{3} - \frac{5}{{11}} + \frac{1}{4}} \right):\left( {1 + \frac{5}{{12}} - \frac{7}{{11}}} \right)\)

\(\begin{array}{l} = \left( {\frac{{2.11.4}}{{3.11.4}} - \frac{{5.3.4}}{{11.3.4}} + \frac{{1.3.11}}{{4.3.11}}} \right):\left( {\frac{11.12}{11.12} + \frac{{5.11}}{{12.11}} - \frac{{7.12}}{{11.12}}} \right)\\ = \left( {\frac{{88 - 60 + 33}}{{121}}} \right):\left( { \frac{{121+55 - 84}}{{121}}} \right)\\ = \frac{{61}}{{121}}:\frac{{92}}{{121}} = \frac{{61}}{{121}}.\frac{{121}}{{92}}= \frac{{61}}{{92}}\end{array}\)

c) \(\left( {13,6 - 37,8} \right).\left( { - 3,2} \right)\)

\( = \left( { - 24,2} \right).\left( { - 3,2} \right) = 77,44\)

d) \(\left( { - 25,4} \right).\left( {18,5 + 43,6 - 16,8} \right):12,7\)

\(\begin{array}{l} = \left( { - 25,4} \right).\left( {62,1 - 16,8} \right):12,7\\ = \left( { - 25,4} \right).45,3:12,7\\ = \left( { - 25,4} \right):12,7.45,3\\ =  (- 2).45,3 =  - 90,6\end{array}\)

a: \(=\dfrac{2}{5}\cdot\left(-\dfrac{3}{7}-\dfrac{5}{14}\right)-\dfrac{18}{35}\)

\(=\dfrac{2}{5}\cdot\dfrac{-6-5}{14}-\dfrac{18}{35}\)

\(=\dfrac{2}{5}\cdot\dfrac{-11}{14}-\dfrac{18}{35}=-\dfrac{22}{70}-\dfrac{18}{35}=\dfrac{-58}{70}=-\dfrac{29}{35}\)

b: \(=\dfrac{88-60+33}{132}:\dfrac{132+55-84}{132}\)

\(=\dfrac{61}{132}\cdot\dfrac{132}{103}=\dfrac{61}{103}\)

c: \(=-24.2\cdot\left(-3.2\right)=24.2\cdot3.2=77.44\)

d: \(=\dfrac{-25.4}{12.7}\cdot45.3=-2\cdot45.3=-90.6\)

\(\frac{1}{2}-\frac{2}{3}+\frac{3}{4}-\frac{4}{5}-\frac{6}{7}-\frac{5}{6}+\frac{4}{5}-\frac{3}{4}+\frac{2}{3}-\frac{1}{2}\)

\(=\left(\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{2}{3}-\frac{2}{3}\right)+\left(\frac{3}{4}-\frac{3}{4}\right)-\left(\frac{4}{5}-\frac{4}{5}\right)-\left(\frac{6}{7}+\frac{5}{6}\right)\)

\(=-\frac{71}{42}\)

71/42

\(\frac{\frac{2}{5}+\frac{2}{7}-\frac{2}{11}}{\frac{3}{5}+\frac{3}{7}-\frac{3}{11}}+\frac{\frac{1}{4}-\frac{1}{5}+\frac{1}{7}}{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}}\)

\(=\frac{2\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}{3\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}+\frac{1\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}\right)}{3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}\right)}\)

\(=\frac{2}{3}+\frac{1}{3}\)

\(=1\)