Dễ

\(\frac{2}{5}+\frac{3}{5}\times\left(3x-3,7\right)=-\frac{53}{10}\)

\(\frac{3}{5}\times\left(3x-3,7\right)=-\frac{53}{10}-\frac{2}{5}\)

\(\frac{3}{5}\times\left(3x-3,7\right)=-\frac{57}{10}\)

\(\left(3x-3,7\right)=-\frac{57}{10}\div\frac{3}{5}\)

\(3x-3,7=-\frac{19}{2}\)

\(3x=-\frac{19}{2}+3,7\)

\(3x=-\frac{29}{5}\)

\(x=-\frac{29}{5}\div3\)

\(x=-\frac{29}{15}\)

Chúc bạn học tốt ^^

3/5.(3x-37/10)= -53/10-2/5

3/5.(3x-37/10)=-57/10

3x-37/10 =-57/10 : 3/5

3x-37/10=-19/2

3x =-19/2+37/10

3x =-29/15

x=-29/15 : 3

x=-29/15

x : 2/5 + 3/5 . (3x - 3,7) = -53/10

x. 2/5 + 3/5. 3x - 3/5. 37/10 = -53/10

x . 5/2 + 9/5.x - 111/50 = -53/10

x. ( 5/2 + 9/5) = -53/10 + 111/50

x . 43/10 = -73/25

x = -73/25 : 43/10

x = -146/215

\(\frac{2}{5}+\frac{3}{5}\times\left(3x-3,7\right)=-\frac{53}{10}\)

\(\frac{3}{5}\times\left(3x-3,7\right)=-\frac{53}{10}-\frac{2}{5}\)

\(\frac{3}{5}\times\left(3x-3,7\right)=\frac{-53-4}{10}\)

\(\frac{3}{5}\times\left(3x-3,7\right)=-\frac{57}{10}\)

\(3x-3,7=-\frac{57}{10}\div\frac{3}{5}\)

\(3x-3,7=-\frac{57}{10}\times\frac{5}{3}\)

\(3x-3,7=-\frac{19}{2}\)

\(3x=-\frac{19}{2}+\frac{37}{10}\)

\(3x=\frac{-95+37}{10}\)

\(3x=-\frac{58}{10}\)

\(x=-\frac{29}{5}\div3\)

\(x=-\frac{29}{5}\times\frac{1}{3}\)

\(x=-\frac{29}{15}\)

Chúc bạn học tốt

Mà ban nè -29/5 ở mo ra rứa bạn

\(-\frac{29}{5}\) là rút gọn của \(-\frac{58}{10}\) đó bn

a) \(x=-\frac{1}{2}\)

b) \(x=-\frac{29}{15}\)

c) \(x=\frac{5}{6}\)

**** cho mình nhé

a) -1/2

b) -29/15

c) 5/6

****

a) \(\frac{2}{5}+\frac{3}{5}.\left(3x-3,7\right)=\frac{-53}{10}\)

=> \(\frac{3}{5}.\left(3x-3,7\right)=\frac{-53}{10}-\frac{2}{5}\)

=> \(\frac{3}{5}.\left(3x-3,7\right)=\frac{-57}{10}\)

=> \(\left(3x-3,7\right)=\frac{-19}{5}\)

=>\(3x=\frac{-19}{5}+\frac{37}{10}\)

=>\(3x=\frac{-1}{10}\)

=>\(x=\frac{-1}{30}\)

Ko biết đúng hay sai!!?

4.x=29/12;-29/12

Câu cuối ban viết rõ lên nhé

a) \(\left(\dfrac{3}{5}+0,415-\dfrac{3}{200}\right).2\dfrac{2}{3}.0,25\)

\(=\left(\dfrac{3}{5}+\dfrac{83}{200}-\dfrac{3}{200}\right).\dfrac{8}{3}.\dfrac{1}{4}\)

\(=\left(\dfrac{3}{5}+\dfrac{2}{5}\right).\dfrac{2}{3}\)

\(=1.\dfrac{2}{3}=\dfrac{2}{3}\)

b) \(\dfrac{2}{5}+\dfrac{3}{5}.\left(3x-3,7\right)=\dfrac{-53}{10}\)

\(\dfrac{3}{5}.\left(3x-\dfrac{37}{10}\right)=\dfrac{-53}{10}-\dfrac{2}{5}=\dfrac{-57}{10}\)

\(3x-\dfrac{37}{10}=\dfrac{-57}{10}:\dfrac{3}{5}=\dfrac{-19}{2}\)

\(3x=\dfrac{-19}{2}+\dfrac{37}{10}=\dfrac{-29}{5}\)

\(x=\dfrac{-29}{5}:3=\dfrac{-29}{15}\)