\(\frac{4}{5}\)+ \(\frac{1}{5}\): \(\frac{-2}{15}\)= \(\frac{4}{5}\)+ \(\frac{1}{5}\)* \(\frac{-15}{2}\)=\(\frac{4}{5}\) + \(\frac{-3}{2}\)= \(\frac{-7}{10}\)

b = \(\frac{-7}{12}\). (\(\frac{3}{11}\)+ \(\frac{-14}{11}\)) + \(\frac{5}{6}\)= \(\frac{-7}{12}\). (-1) + \(\frac{5}{6}\)= \(\frac{7}{12}\)+ \(\frac{10}{12}\)= \(\frac{17}{12}\)

x-\(\frac{3}{5}\)x = \(\frac{2}{5}\)

=> x.(1-\(\frac{3}{5}\)) =\(\frac{2}{5}\)

=> x. \(\frac{2}{5}\)= \(\frac{2}{5}\)

=> x=1

Vậy x=1

[6: 3/5 - 7/6 x 6/7] : [ 21/5 x 10/11 + 57/11]

= 9 : 9

= 1 

\(\left(6:\frac{3}{5}-\frac{7}{6}\times\frac{6}{7}\right):\left(\frac{21}{5}\times\frac{10}{11}+\frac{57}{11}\right)\)

\(=\left(10-1\right):\left(\frac{42}{11}+\frac{57}{11}\right)\)

\(=9:9\)

\(=1\)

\(\left(6\div\frac{3}{5}-\frac{7}{6}\times\frac{6}{7}\right)\div\left(\frac{21}{5}\times\frac{10}{11}+\frac{57}{11}\right)\)

\(=\left(6\times\frac{5}{3}-1\right)\div\left(\frac{42}{11}+\frac{57}{11}\right)\)

\(=\frac{13}{5}\div9\)

\(=\frac{13}{45}\)

a= -1/7

b -50/21

c   1/8

d 19/56

e-13/24

f -89/312

\(\left(6:3,5-1\dfrac{1}{6}\times\dfrac{6}{7}\right):\left(4,2\times\dfrac{10}{11}+5\dfrac{2}{11}\right)\\ =\left(\dfrac{12}{7}-\dfrac{7}{6}\times\dfrac{6}{7}\right):\left(\dfrac{21}{5}\times\dfrac{10}{11}+\dfrac{57}{11}\right)\\ =\left(\dfrac{17}{7}-1\right):\left(\dfrac{42}{11}+\dfrac{57}{11}\right)\\ =\dfrac{10}{7}:9\\ =\dfrac{10}{63}\)

\(\left(6:\dfrac{3}{5}-1\dfrac{1}{6}\times\dfrac{6}{7}\right):\left(4,2\times\dfrac{10}{11}+5\dfrac{2}{11}\right)\)

\(=\left(6\times\dfrac{5}{3}-\dfrac{7}{6}\times\dfrac{6}{7}\right):\left(\dfrac{21}{5}\times\dfrac{10}{11}+\dfrac{57}{11}\right)\)

\(=\left(10-1\right):\left(\dfrac{42}{11}+\dfrac{57}{11}\right)=9:9=1\)

Một hình thang có hiệu hai đáy là 18 cm, đáy bé bằng 3/5 đáy lớn, đáy lớn bằng 3/2 chiều cao. Tình diện tích của hình thang đó ?

Hỡi các thiên tài toán học, hãy giải giúp mình bài toán này !

                                                                                Trân trọng cảm ơn !

a) \(2.\left(x+\frac{2}{5}\right)+1\frac{1}{4}=\frac{11}{20}\)

\(2.\left(x+\frac{2}{5}\right)+\frac{5}{4}=\frac{11}{20}\)

\(2.\left(x+\frac{2}{5}\right)=\frac{-7}{10}\)

\(x+\frac{2}{5}=\frac{-7}{20}\)

\(x=\frac{-13}{20}\)

Vậy \(x=\frac{-13}{20}\)

b)\(x-1\frac{1}{8}-\frac{2}{3}x-\frac{5}{6}x=75\%\)

  \(\left(x-\frac{2}{3}x-\frac{5}{6}x\right)-\frac{9}{8}=\frac{3}{4}\)

\(\frac{-1}{2}x-\frac{9}{8}=\frac{3}{4}\)

\(\frac{-1}{2}x=\frac{15}{8}\)

\(x=\frac{-15}{4}\)

Vậy \(x=\frac{-15}{4}\)

\(a,2.\left(x+\frac{2}{5}\right)+1\frac{1}{4}=\frac{11}{20}\)

\(2.\left(x+\frac{2}{5}\right)+\frac{5}{4}=\frac{11}{20}\)

\(2.\left(x+\frac{2}{5}\right)=\frac{11}{20}-\frac{5}{4}\)

\(2.\left(x+\frac{2}{5}\right)=\frac{-7}{10}\)

\(x+\frac{2}{5}=\frac{-7}{10}:2\)

\(x+\frac{2}{5}=\frac{-7}{20}\)

\(x=\frac{-7}{20}-\frac{2}{5}\)

\(x=\frac{-3}{4}\)