27/15 + 6/8 =51/20
19/24 + 7/18 = 85/72
2/9 - 1/6 = 1/18
8/15 -1/3 = 1/5

a: \(1-\left(5\dfrac{4}{9}+a-7\dfrac{7}{18}\right):15\dfrac{3}{4}=0\)

=>\(\left(5+\dfrac{4}{9}+a-7-\dfrac{7}{18}\right):\dfrac{63}{4}=1\)

=>\(\left(a-2+\dfrac{1}{18}\right)=\dfrac{63}{4}\)

=>\(a-\dfrac{35}{18}=\dfrac{63}{4}\)

=>\(a=\dfrac{63}{4}+\dfrac{35}{18}=\dfrac{637}{36}\)

b: \(B=\left(\dfrac{2}{15}+\dfrac{5}{3}-\dfrac{3}{5}\right):\left(4\dfrac{2}{3}-2\dfrac{1}{2}\right)\)

\(=\dfrac{2+5\cdot5-3^2}{15}:\left(4+\dfrac{2}{3}-2-\dfrac{1}{2}\right)\)

\(=\dfrac{2+4^2}{15}:\left(2+\dfrac{2}{3}-\dfrac{1}{2}\right)\)

\(=\dfrac{18}{15}:\dfrac{13}{6}=\dfrac{6}{5}\cdot\dfrac{6}{13}=\dfrac{36}{65}\)

\(\dfrac{3}{5}+\dfrac{1}{2}+\dfrac{8}{15}\\ =\dfrac{3\times6}{5\times6}+\dfrac{1\times15}{2\times15}+\dfrac{8\times2}{15\times2}\\ =\dfrac{18}{30}+\dfrac{15}{30}+\dfrac{16}{30}\\ =\dfrac{49}{30}\\ \dfrac{6}{9}+\dfrac{14}{18}-\dfrac{5}{6}\\ =\dfrac{6\times2}{9\times2}+\dfrac{14}{18}-\dfrac{5\times3}{6\times3}\\ =\dfrac{12}{18}+\dfrac{14}{18}-\dfrac{15}{18}\\ =\dfrac{11}{18}\)

\(\dfrac{9}{20}-\dfrac{3}{5}:\dfrac{4}{1}\\ =\dfrac{9}{20}-\dfrac{3}{5}\times\dfrac{1}{4}\\ =\dfrac{9}{20}-\dfrac{3}{20}\\ =\dfrac{6}{20}\\ =\dfrac{3}{10}\)

\(\dfrac{1}{6}+\dfrac{2}{3}\times\dfrac{8}{9}\\=\dfrac{1}{6}+\dfrac{16}{27}\\ =\dfrac{1\times9}{6\times9}+\dfrac{16\times2}{27\times2}\\ =\dfrac{9}{54}+\dfrac{32}{54}\\ =\dfrac{41}{54}.\)

a) Ta có: \(\dfrac{-5}{18}+\dfrac{32}{45}-\dfrac{9}{10}\)

\(=\dfrac{-25}{90}+\dfrac{64}{90}-\dfrac{81}{90}\)

\(=\dfrac{-42}{90}=-\dfrac{7}{15}\)

b) Ta có: \(\left(-\dfrac{1}{4}+\dfrac{51}{33}-\dfrac{5}{3}\right)-\left(-\dfrac{15}{12}+\dfrac{6}{11}-\dfrac{42}{29}\right)\)

\(=\dfrac{-1}{4}+\dfrac{17}{11}-\dfrac{5}{3}+\dfrac{5}{4}-\dfrac{6}{11}+\dfrac{42}{29}\)

\(=\dfrac{-5}{3}+\dfrac{42}{29}\)

\(=\dfrac{-145}{87}+\dfrac{126}{87}=\dfrac{-19}{87}\)

c) Ta có: \(1-\dfrac{1}{2}+2-\dfrac{2}{3}+3-\dfrac{3}{4}+4-\dfrac{1}{4}-3-\dfrac{1}{3}-2-\dfrac{1}{2}-1\)

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

\(=-1-1-1+4\)

=1

a) Ta có: 

b) Ta có: 

c) Ta có: 

=1

g)\(=\left(-\dfrac{3}{4}+\dfrac{2}{5}\right).\dfrac{7}{3}+\left(\dfrac{3}{5}+-\dfrac{1}{4}\right).\dfrac{7}{3}\)
   \(=\left(-\dfrac{3}{4}+-\dfrac{1}{4}+\dfrac{2}{5}+\dfrac{3}{5}\right).\dfrac{7}{3}\)
\(=\left(-1+1\right).\dfrac{7}{3}=0.\dfrac{7}{3}=0\)

f) \(\dfrac{15^3+5.15^2-5^3}{18^3+6.18^2-6^3}\)

\(=\dfrac{3^3.5^3+5.5^2.3^2-5^3}{3^3.6^3+6.6^2.3^2-6^3}\)

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

\(=\dfrac{5^3}{6^3}\)

\(=\dfrac{125}{216}\)

\(\dfrac{15}{12}+\dfrac{5}{13}-\dfrac{3}{12}-\dfrac{18}{13}-\dfrac{1}{3}\)

\(=\left(\dfrac{15}{12}-\dfrac{3}{12}\right)+\left(\dfrac{5}{13}-\dfrac{18}{13}\right)-\dfrac{1}{3}\)

\(=-1+1-\dfrac{1}{3}\)

\(=0-\dfrac{1}{3}\)

\(=\dfrac{-1}{3}\)

------------------------------------------

\(14.\dfrac{3}{2}+\dfrac{6}{5}:\left(-\dfrac{2}{5}\right)\)

\(=14.\dfrac{3}{2}+\dfrac{6}{5}.\dfrac{-5}{2}\)

\(=21+\dfrac{6}{5}.\dfrac{-5}{2}\)

\(=21+\left(-3\right)\)

\(=18\)

------------------------------------------------

\(\sqrt{\dfrac{1}{4}+\dfrac{2}{3}-\left(\dfrac{1}{3}\right)^2}\)

\(=\sqrt{\dfrac{1}{4}+\dfrac{2}{3}-\dfrac{1}{9}}\)

\(=\sqrt{\dfrac{3}{12}+\dfrac{8}{12}-\dfrac{1}{9}}\)

\(=\sqrt{\dfrac{11}{12}-\dfrac{1}{9}}\)

\(=\sqrt{\dfrac{99}{108}-\dfrac{12}{108}}\)

\(=\sqrt{\dfrac{29}{36}}\)

\(=\dfrac{\sqrt{29}}{6}\)

\(\dfrac{15}{12}+\dfrac{5}{13}-\dfrac{3}{12}-\dfrac{18}{13}-\dfrac{1}{3}\)
\(=\dfrac{5}{4}+\dfrac{5}{13}-\dfrac{1}{4}-\dfrac{18}{13}-\dfrac{1}{3}\)
\(=\left(\dfrac{5}{4}-\dfrac{1}{4}\right)+\left(\dfrac{5}{13}-\dfrac{18}{13}\right)-\dfrac{1}{3}\)
\(=1+\left(-1\right)-\dfrac{1}{3}=0-\dfrac{1}{3}=-\dfrac{1}{3}\)

\(a.\)

\(-\dfrac{2}{9}+\dfrac{7}{-9}=\dfrac{-2}{9}-\dfrac{7}{9}=\dfrac{-2-7}{9}=\dfrac{-9}{9}=-1\)

\(b.\)

\(\dfrac{1}{-6}+\dfrac{13}{-15}=-\dfrac{1}{6}-\dfrac{13}{15}=\dfrac{\left(-1\right)\cdot5-13\cdot2}{30}=\dfrac{-5-26}{30}=\dfrac{-31}{30}\)

\(c.\)

\(\dfrac{5}{-6}+\dfrac{-5}{12}+\dfrac{7}{18}=\dfrac{-5}{6}-\dfrac{5}{12}+\dfrac{7}{18}=\dfrac{\left(-5\right)\cdot6-5\cdot3+7\cdot2}{36}=\dfrac{-31}{36}\)

a) \(\dfrac{-2}{9}+\dfrac{7}{-9}=\dfrac{-2+-7}{9}=-1\) 

b)\(\dfrac{1}{-6}+\dfrac{13}{-15}=\dfrac{-5}{30}+\dfrac{26}{30}=\dfrac{-5+26}{30}=\dfrac{7}{10}\) 

c) \(\dfrac{5}{-6}+\dfrac{-5}{12}+\dfrac{7}{18}=\dfrac{-30}{36}+\dfrac{-15}{36}+\dfrac{18}{36}=\dfrac{-30+-15+18}{36}=\dfrac{-3}{4}\)

Đề bài là:Tính các giá trị biểu thức sau ạ

a: \(=\left(9-\dfrac{13}{18}\right):\dfrac{325}{27}-\dfrac{17}{8}:\dfrac{51}{40}\)

\(=\dfrac{149}{18}\cdot\dfrac{27}{325}-\dfrac{17}{8}\cdot\dfrac{40}{51}\)

\(=\dfrac{447}{650}-\dfrac{5}{3}=-\dfrac{1909}{1950}\)

b: \(=\dfrac{48}{64}+\left(\dfrac{4}{5}-2-\dfrac{4}{15}\right):\dfrac{11}{3}\)

\(=\dfrac{3}{4}+\dfrac{-22}{15}\cdot\dfrac{3}{11}=\dfrac{3}{4}-\dfrac{2}{5}=\dfrac{15-8}{20}=\dfrac{7}{20}\)

1) \(\dfrac{1}{2}+\dfrac{13}{19}-\dfrac{4}{9}+\dfrac{6}{19}+\dfrac{5}{18}\)

\(=\dfrac{1}{2}+\left(\dfrac{13}{19}+\dfrac{6}{19}\right)-\dfrac{4}{9}+\dfrac{5}{18}\)

\(=\dfrac{3}{2}-\dfrac{4}{9}+\dfrac{5}{18}\)

\(=\dfrac{19}{18}+\dfrac{5}{18}\)

\(=\dfrac{24}{18}\)

\(=\dfrac{4}{3}\)

2) \(\dfrac{-20}{23}+\dfrac{2}{3}-\dfrac{3}{23}+\dfrac{2}{5}+\dfrac{7}{15}\)

\(=\left(-\dfrac{20}{23}-\dfrac{3}{23}\right)+\dfrac{2}{3}+\dfrac{2}{5}+\dfrac{7}{15}\)

\(=-1+\dfrac{2}{3}+\dfrac{2}{5}+\dfrac{7}{15}\)

\(=-\dfrac{1}{3}+\dfrac{2}{5}+\dfrac{7}{15}\)

\(=\dfrac{1}{15}+\dfrac{7}{15}\)

\(=\dfrac{8}{15}\)

3) \(\dfrac{4}{3}+\dfrac{-11}{31}+\dfrac{3}{10}-\dfrac{20}{31}-\dfrac{2}{5}\)

\(=\left(\dfrac{-11}{31}-\dfrac{20}{31}\right)+\dfrac{4}{3}+\dfrac{3}{10}-\dfrac{2}{5}\)

\(=-1+\dfrac{4}{3}+\dfrac{3}{10}-\dfrac{2}{5}\)

\(=\dfrac{1}{3}+\dfrac{3}{10}-\dfrac{2}{5}\)

\(=\dfrac{1}{3}-\dfrac{1}{10}\)

\(=\dfrac{7}{30}\)

4) \(\dfrac{5}{7}.\dfrac{5}{11}+\dfrac{5}{7}.\dfrac{2}{11}-\dfrac{5}{7}.\dfrac{14}{11}\)

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

\(=\dfrac{5}{7}.-\dfrac{7}{11}\)

\(=-\dfrac{35}{77}\)

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