a) \(\dfrac{7}{30}+\dfrac{\left(-12\right)}{37}+\dfrac{23}{30}+\dfrac{\left(-25\right)}{37}=\left(\dfrac{7}{30}+\dfrac{23}{30}\right)+\left(\dfrac{-12}{37}+\dfrac{-25}{37}\right)=1+\left(-1\right)=0\)

b) \(\dfrac{5}{7}\cdot\dfrac{5}{11}+\dfrac{5}{7}\cdot\dfrac{2}{11}-\dfrac{5}{7}\cdot\dfrac{14}{11}=\dfrac{5}{7}\cdot\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)=\dfrac{5}{7}\cdot\left(-\dfrac{7}{11}\right)=-\dfrac{5}{11}\)

c) \(\dfrac{\left(-5\right)}{7}\cdot\dfrac{3}{13}-\dfrac{5}{7}\cdot\dfrac{10}{13}+1\dfrac{5}{7}=\dfrac{5}{7}\cdot\dfrac{-3}{13}-\dfrac{5}{7}\cdot\dfrac{10}{13}+\dfrac{12}{7}=\dfrac{5}{7}\cdot\left(\dfrac{-3}{13}-\dfrac{10}{13}\right)+\dfrac{12}{7}=\dfrac{5}{7}\cdot\left(-1\right)+\dfrac{12}{7}=\left(-\dfrac{5}{7}\right)+\dfrac{12}{7}=\dfrac{7}{7}=1\)

b: =12+5/14-3-5/7-5-5/14

=4-5/7

=28/7-5/7=23/7

c: =(-2/5-11/10)+(7/11-7/11)

=-4/10-11/10=-15/10=-3/2

\(a,\dfrac{5}{9}\cdot\dfrac{7}{13}+\dfrac{5}{9}\cdot\dfrac{8}{13}-\dfrac{5}{13}\cdot\dfrac{2}{9}\)

\(=\dfrac{5}{9}\cdot\dfrac{7}{13}+\dfrac{5}{9}\cdot\dfrac{8}{13}-\dfrac{2}{13}\cdot\dfrac{5}{9}\)

\(=\dfrac{5}{9}\cdot\left(\dfrac{7}{13}+\dfrac{8}{13}-\dfrac{2}{13}\right)\)

\(=\dfrac{5}{9}\cdot\dfrac{14}{13}\)

\(=\dfrac{70}{117}\)

\(d,\dfrac{1}{2}+\dfrac{-2}{3}+\dfrac{1}{6}+\dfrac{-2}{5}\)

\(=\left(\dfrac{1}{2}+\dfrac{-2}{3}+\dfrac{1}{6}\right)+\dfrac{-2}{5}\)

\(=0+\dfrac{-2}{5}\)

\(=\dfrac{-2}{5}\)

-10/13 + 5/17 - 3/13 + 12/17 - 11/20                                              3/4 + -5/6 - 11/-12

(-10/13 - 3/13) + (5/17 + 12/17) - 11/20                                         3/4 - 5/6 - 11/-12

-13/13 + 17/17 - 11/20                                                                   9/12 - 10/12 - 1/-12

-1 + 1 - 11/20                                                                                 -1/12 - -1/12

=-11/20                                                                                           -1/12 + 1/12

                                                                                                        =0

                              k cho mk nhé   cảm ơn bạn rất nhiều

a) \(\dfrac{-5}{9}-\dfrac{-5}{12}=\dfrac{-5}{9}+\dfrac{5}{12}=\dfrac{-20}{36}+\dfrac{15}{36}=-\dfrac{5}{36}\)

b) \(\dfrac{-5}{12}:\dfrac{15}{4}=\dfrac{-5}{12}\times\dfrac{4}{15}=\dfrac{-1}{9}\)

c) \(\dfrac{1}{13}\cdot\dfrac{8}{13}+\dfrac{5}{13}\cdot\dfrac{1}{13}-\dfrac{14}{13}=\dfrac{1}{13}\cdot\left(\dfrac{8}{13}+\dfrac{5}{13}\right)-\dfrac{14}{13}=\dfrac{1}{13}\cdot1-\dfrac{14}{13}=\dfrac{1}{13}-\dfrac{14}{13}=-1\)

a)=--5/36

b)=-1/9

c) =-1 

cách làm như trên

a: Ta có: \(\left(8\cdot5^7+5^6-5^5\right):5^5\)

\(=8\cdot5^2+5-1\)

\(=200+4=204\)

b: Ta có: \(\left(9^{30}-27^{19}\right):3^{57}+\left(125^9-25^{12}\right):5^{24}\)

\(=3^{60}:3^{57}-3^{57}:3^{57}+5^{27}:5^{24}-5^{24}:5^{24}\)

\(=27-1+125-1\)

=150

a. (8,5⁷ - 5⁵ + 5⁶) : 5⁵

= (8,5⁷ : 5⁵) - (5⁵ : 5⁵) + (5⁶ : 5⁵)

= 1,7² - 1 + 5

= 2,89 - 1 + 5

= 6,89

b. (9³⁰ - 27¹⁹) : 3⁵⁷ + (125⁹ - 25¹²) : 5²⁴

= (9³⁰ : 3⁵⁷) - (27¹⁹ : 3⁵⁷) + (125⁹ : 5²⁴) - (25¹² : 5²⁴)

= 3³ - 1 + 125 - 1

= 27 - 1 + 125 - 1

= 150

c. (10¹² + 5¹¹ . 2⁹ - 5¹³ - 2⁸) : 4 . 5⁵ . 10⁶

= (10¹² + 2,5 , 10¹⁰ - 5¹³ - 2⁸) : 1,25 . 10¹⁰

= (10¹² : 1,25 . 10¹⁰) + (2,5 . 10¹⁰ : 1,25 . 10¹⁰) - (5¹³ : 1,25 . 10¹⁰) - (2⁸ : 1,25 . 10¹⁰)

= 80 + 2 - \(\dfrac{25}{256}\) - \(\dfrac{1}{48828125}\)

= 81,90234373 \(\approx\) 82

a; - \(\dfrac{10}{13}\) + \(\dfrac{5}{17}\) - \(\dfrac{3}{13}\) + \(\dfrac{12}{17}\) - \(\dfrac{11}{20}\)

= - (\(\dfrac{10}{13}\) + \(\dfrac{3}{13}\)) + (\(\dfrac{5}{17}\) + \(\dfrac{12}{17}\)) - \(\dfrac{11}{20}\)

= - 1 + 1  - \(\dfrac{11}{20}\)

=   0 - \(\dfrac{11}{20}\)

= - \(\dfrac{11}{20}\)

b; \(\dfrac{3}{4}\) + \(\dfrac{-5}{6}\) - \(\dfrac{11}{-12}\)

= \(\dfrac{9}{12}\) - \(\dfrac{10}{12}\) + \(\dfrac{11}{12}\)

= \(\dfrac{10}{12}\)

= \(\dfrac{5}{6}\)

c; [13.\(\dfrac{4}{9}\) + 2.\(\dfrac{1}{9}\)] - 3.\(\dfrac{4}{9}\)

= [\(\dfrac{52}{9}\) + \(\dfrac{2}{9}\)] - \(\dfrac{4}{3}\)

= \(\dfrac{54}{9}\) - \(\dfrac{4}{3}\)

= \(\dfrac{14}{3}\)

d; 1,25 : \(\dfrac{15}{20}\) + [25% - \(\dfrac{5}{6}\)] : 4.\(\dfrac{2}{3}\)

= 1,25  x \(\dfrac{20}{15}\) + [\(\dfrac{1}{4}\) - \(\dfrac{5}{6}\)] : 4.\(\dfrac{2}{3}\)

=  \(\dfrac{5}{3}\) - \(\dfrac{7}{12}\) : 4 x \(\dfrac{2}{3}\)

= \(\dfrac{5}{3}\) - \(\dfrac{7}{48}\).\(\dfrac{2}{3}\)

= \(\dfrac{5}{3}\) - \(\dfrac{7}{72}\)

= \(\dfrac{113}{72}\)

\(\left(10^{12}+5^{11}.2^9-5^{13}.2^8\right):4.5^5.10^6\)

\(=\left(5^{12}.2^{12}+5^{11}.2^9-5^{13}.2^8\right):2^2.5^5.2^6.5^6\)

\(=\left[5^{11}.2^8\left(5.2^4+2-5^2\right)\right]:2^8.5^{11}\)

\(=\frac{5^{11}.2^8\left(5.16+2-25\right)}{5^{11}.2^8}\)

\(=80+2-25\)

\(=57\)

\(A=\frac{155-\frac{10}{7}-\frac{5}{11}+\frac{5}{23}}{403-\frac{26}{7}-\frac{13}{11}+\frac{13}{23}}-\frac{\frac{3}{5}+\frac{3}{13}-0,9}{\frac{7}{91}+0,2-\frac{3}{10}}\)

\(A=\frac{155-5\left(\frac{2}{7}-\frac{1}{11}+\frac{1}{23}\right)}{403-13\left(\frac{2}{7}-\frac{1}{11}+\frac{1}{23}\right)}-\frac{\frac{3}{5}+\frac{3}{13}-\frac{9}{10}}{\frac{7}{91}+\frac{2}{10}-\frac{3}{10}}\)

\(A=\frac{155-5}{403-13}-\frac{3\left(\frac{1}{5}+\frac{1}{13}\right)-\frac{9}{10}}{\frac{7}{91}+\left(-\frac{1}{10}\right)}\)

\(A=\frac{5}{13}-\frac{\left(-\frac{9}{130}\right)}{\left(-\frac{3}{130}\right)}=\frac{5}{13}-\frac{\frac{9}{130}}{\frac{3}{130}}\)

\(A=\frac{5}{13}-\frac{9}{130}\cdot\frac{130}{3}\)

\(A=\frac{5}{13}-3=-\frac{34}{13}\)

\(B=\frac{30\cdot4^7\cdot3^{29}-5\cdot14^5\cdot2^{12}}{54\cdot6^{14}\cdot9^7-12\cdot8^5\cdot7^5}\)

\(B=\frac{30\cdot\left(2^2\right)^7\cdot3^{29}-5\cdot\left(2\cdot7\right)^5\cdot2^{12}}{54\cdot\left(2\cdot3\right)^{14}\cdot\left(3^2\right)^7-12\cdot\left(2^3\right)^5\cdot7^5}\)

\(B=\frac{30\cdot2^{14}\cdot3^{29}-5\cdot2^5\cdot7^5\cdot2^{12}}{54\cdot2^{14}\cdot3^{14}\cdot3^{14}-12\cdot2^{15}\cdot7^5}\)

\(B=\frac{30\cdot3^{29}-5\cdot2^{17}\cdot7^5}{54\cdot3^{28}-12\cdot2^{15}\cdot7^5}=\frac{30\cdot3-5\cdot2^2}{54-12}=\frac{5}{3}\)