\(\dfrac{15}{34}+\dfrac{7}{21}+\dfrac{19}{34}-\dfrac{32}{17}+\dfrac{2}{3}\) 

\(=\left(\dfrac{15}{34}+\dfrac{19}{34}\right)+\dfrac{7}{21}+\dfrac{-32}{17}+\dfrac{2}{3}\) 

\(=1+\dfrac{1}{3}+\dfrac{-32}{17}+\dfrac{2}{3}\) 

\(=1+\left(\dfrac{1}{3}+\dfrac{2}{3}\right)+\dfrac{-32}{17}\) 

\(=1+1+\dfrac{-32}{17}\) 

\(2+\dfrac{-32}{17}=\dfrac{2}{17}\)

x=-36

x=11

x=100

x=14

x=-36

x=11

x=100

x=14

a, >

b, <

c, <

Câu 1:

\(=\dfrac{15}{34}+\dfrac{19}{34}-1-\dfrac{15}{17}+\dfrac{1}{3}+\dfrac{3}{5}\)

\(=-\dfrac{15}{17}+\dfrac{14}{15}=\dfrac{13}{255}\)

Câu 2: 

\(=\dfrac{5^4\cdot5^4\cdot2^8}{4^4\cdot6^4\cdot3^2\cdot5}=\dfrac{5^7}{6^4\cdot3^2}\)

a, \(\dfrac{15}{34}+\dfrac{7}{21}+\dfrac{19}{34}-\dfrac{20}{15}+\dfrac{3}{7}\)

= \(\left(\dfrac{15}{34}+\dfrac{19}{34}\right)+\left(\dfrac{7}{21}+\dfrac{3}{7}\right)-\dfrac{20}{15}\)

= 1 + \(\dfrac{16}{21}-\dfrac{20}{15}\)

= \(\dfrac{3}{7}\)

b, \(\dfrac{27}{25}+\dfrac{4}{21}-\dfrac{2}{25}+\dfrac{17}{21}-\dfrac{1}{2}\)

= \(\left[\dfrac{27}{25}+\left(-\dfrac{2}{25}\right)\right]+\left(\dfrac{4}{21}-\dfrac{7}{21}\right)-\dfrac{1}{2}\)

= 1 + \(\left(\dfrac{-1}{7}\right)-\dfrac{1}{2}\) = \(\dfrac{5}{14}\)

\(\dfrac{x-17}{33}+\dfrac{x-21}{29}+\dfrac{x}{25}=4\)

\(\dfrac{x-17}{33}+\dfrac{x-21}{29}+\dfrac{x}{25}-4=0\)

\(\dfrac{\left(x-17\right)\times725}{33\times725}+\dfrac{\left(x-21\right)\times825}{29\times825}+\dfrac{x\times957}{25\times957}-\dfrac{4\times23925}{23925}=0\)

\(725x-12325+825x-17325+957x-95700=0\)

\(2507x-125350=0\)

\(2507x=125350\)

\(x=50\)

Nếu mà theo cách x - 50 = 0 thì bạn theo cách này nha:

\(\dfrac{x-17}{33}+\dfrac{x-21}{29}+\dfrac{x}{25}=4\)

\(\dfrac{x-17}{33}+\dfrac{x-21}{29}+\dfrac{x}{25}-4=0\)

\(\dfrac{x-17}{33}-1+\dfrac{x-21}{29}-1+\dfrac{x}{25}-2=0\)

\(\dfrac{x-50}{33}+\dfrac{x-50}{29}+\dfrac{x-50}{25}=0\)

\(\left(x-50\right)\left(\dfrac{1}{33}+\dfrac{1}{29}+\dfrac{1}{5}\right)=0\)

Vì  \(\dfrac{1}{33}+\dfrac{1}{29}+\dfrac{1}{25}>0\)

=> \(x-50=0\)
=> \(x=50\)