a) \(\dfrac{3}{10}\)

b) \(\dfrac{9}{28}\)

c) \(\dfrac{88}{195}\)

\(B=\dfrac{13}{15}\cdot0,15\cdot3+\left(\dfrac{8}{15}-1\dfrac{19}{60}\right):\left(1\dfrac{23}{24}\right)\)

\(=\dfrac{13}{15}\cdot\dfrac{15}{100}\cdot3+\left(\dfrac{32}{60}-1-\dfrac{19}{60}\right):\dfrac{47}{24}\)

\(=\dfrac{39}{100}+\dfrac{-47}{60}\cdot\dfrac{24}{47}\)

\(=\dfrac{39}{100}-\dfrac{24}{60}=\dfrac{39}{100}-\dfrac{40}{100}=-\dfrac{1}{100}\)

a: =1/6+14/6-3/6=12/6=2

b: =-13/8+5/4:(-5/4)

=-13/8-1=-21/8

c: =-3/8(2/5+14/5)

=-3/8*16/5

=-6/5

d: =5/34(1/4+11/9-2/9+29/4)

=5/34*(15/2+1)

=5/34*17/2
=5/4

a)

\(-12:\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2\)

\(=-12:\left(\dfrac{18}{24}-\dfrac{20}{24}\right)^2\)

\(=-12:\left(\dfrac{-1}{12}\right)^2\)

\(=-12:\dfrac{1}{144}\)

\(=-12\times\dfrac{144}{1}\)

\(=-1728\)

b)

\(\left(2^2:\dfrac{4}{3}-\dfrac{1}{2}\right)\times\dfrac{6}{5}-17\)

\(=\left(4\times\dfrac{3}{4}-\dfrac{1}{2}\right)\times\dfrac{6}{5}-17\)

\(=\left(3-\dfrac{1}{2}\right)\times\dfrac{6}{5}-17\)

\(=\dfrac{5}{2}\times\dfrac{6}{5}-17\)

\(=3-17\)

\(=-14\)

a)\(=-12:\left(-\dfrac{1}{12}\right)^2\)
\(=-12:\dfrac{1}{144}\)\(=-12.144=-1728\)
b)\(=\left(8:\dfrac{4}{3}-\dfrac{1}{2}\right).\dfrac{6}{5}-17\)
\(=\left(6-\dfrac{1}{2}\right).\dfrac{6}{5}-17\)
\(=\dfrac{11}{2}.\dfrac{6}{5}-17=\dfrac{33}{5}-17=\dfrac{33}{5}-\dfrac{85}{5}=-\dfrac{2}{5}\)

   

a. 7/9 - 16/9 = -9/9 = -1

b. 2/-15 + 7/10 = 17/30

c. (4 2/3 - 4 3/4) : -5/12 - 4/5 

= (14/3 - 19/4) : (-5/12) - 4/5

= -1/12 : (-5/12) - 4/5

= 1/5 - 4/5

= -3/5

thanks

Bài 1:

a: \(\sqrt{252}-\sqrt{700}+\sqrt{1008}-\sqrt{448}\)

\(=6\sqrt{7}-10\sqrt{7}+12\sqrt{7}-8\sqrt{7}\)

\(=8\sqrt{7}\)

Bài 3: 

a: \(\sqrt{27^2-23^2}=10\sqrt{2}\)

b: \(\sqrt{37^2-35^2}=12\)

c: \(\sqrt{65^2-63^2}=16\)

d: \(\sqrt{117^2-108^2}=45\)

\(a,27.5^2-25.127=27.25-25.127=25\left(27-127\right)=25.\left(-100\right)=-2500\\ b,\dfrac{-5}{12}+\dfrac{3}{4}+\dfrac{1}{-3}=\dfrac{-5}{12}+\dfrac{9}{12}-\dfrac{4}{12}=\dfrac{0}{12}=0\)

a = -2500

b = 0

bài 3:

\(A=\dfrac{9}{1\cdot2}+\dfrac{9}{2\cdot3}+\dfrac{9}{3\cdot4}+...+\dfrac{9}{2021\cdot2022}\)

\(=9\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{2021\cdot2022}\right)\)

\(=9\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\right)\)

\(=9\cdot\dfrac{2021}{2022}=\dfrac{6063}{674}\)

Bài 1:

a: \(\left(\dfrac{1}{2}+\dfrac{16}{30}\right)-\left(1+\dfrac{1}{30}\right)\)

\(=\dfrac{15+16}{30}-1-\dfrac{1}{30}\)

\(=\dfrac{30}{30}-1\)

=1-1

=0

b: \(\dfrac{-5}{11}\cdot\dfrac{4}{13}+\dfrac{-5}{11}\cdot\dfrac{9}{13}+3\dfrac{5}{11}\)

\(=-\dfrac{5}{11}\left(\dfrac{4}{13}+\dfrac{9}{13}\right)+3+\dfrac{5}{11}\)

\(=-\dfrac{5}{11}+3+\dfrac{5}{11}\)

=3

c: \(3^2-12\left(\dfrac{3}{4}-\dfrac{2}{3}\right)\)

\(=9-12\cdot\dfrac{9-8}{12}\)

=9-1

=8