\(A=\dfrac{-3}{5}.\dfrac{3}{13}+\dfrac{-3}{5}.\dfrac{10}{13}+5\)

\(A=\dfrac{-3}{5}\left(\dfrac{3}{13}+\dfrac{10}{13}\right)+5\)

\(A=\dfrac{-3}{5}.1+5\)

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

\(A=\dfrac{22}{5}\)

cảm ơn bạn nha, làm giúp mk câu b đc ko?

Mấy bài này bạn tự làm đi, chuyển vế tìm x gần giống cấp I mà.

b)\(\dfrac{-3}{5}.x=\dfrac{1}{4}+0,75\)

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

=>\(x=1:\dfrac{-3}{5}\)

=>\(x=\dfrac{-5}{3}\)

Vậy \(x=\dfrac{-5}{3}\)

Ta có: \(S< \dfrac{1}{2}+\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{31}+\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{32}\) \(=\dfrac{1}{2}+\dfrac{3}{11}+\dfrac{2}{31}+\dfrac{2}{32}\)

\(=\dfrac{4909}{5456}< \dfrac{9}{10}\)

\(\Rightarrow S< \dfrac{9}{10}\)

Vậy \(S< \dfrac{9}{10}\)

A = \(\dfrac{17}{15}.\dfrac{-31}{125}.\dfrac{1}{2}.\dfrac{10}{17}.\dfrac{-1}{8}\)

= \(\dfrac{17.\left(-31\right).1.10.\left(-1\right)}{15.125.2.17.8}\)

= \(\dfrac{17.\left[\left(-31\right).\left(-1\right)\right].1.2.5}{5.3.125.17.4.2}\)

= \(\dfrac{31.1}{3.125.4}\)

= \(\dfrac{31}{1500}\)

B = \(\left(\dfrac{11}{4}.\dfrac{-5}{9}-\dfrac{4}{9}.\dfrac{11}{4}\right).\dfrac{8}{33}\)

= \(\left[\dfrac{11}{4}.\left(\dfrac{-5}{9}-\dfrac{4}{9}\right)\right].\dfrac{8}{33}\)

= \(\left(\dfrac{11}{4}.\dfrac{-9}{9}\right).\dfrac{8}{33}\)

= \(\left[\dfrac{11}{4}.\left(-1\right)\right].\dfrac{4.2}{\left(-11\right).\left(-3\right)}\)

= \(\dfrac{-11}{4}.\dfrac{4.2}{\left(-11\right).\left(-3\right)}\)

= \(\dfrac{\left(-11\right).4.2}{4.\left(-11\right)\left(-3\right)}\)

= \(\dfrac{2}{-3}\)

Ok nhá!

\(\dfrac{1}{7}=\dfrac{8}{-x}\)=> \(-x=56\)

=> \(x=56\)

2) => 18x = 18

=> x = 1

3) \(\dfrac{-4}{3}+x=\dfrac{-11}{6}\)

=> \(x=\dfrac{-11}{6}+\dfrac{4}{3}\)

=> \(x=\dfrac{-1}{2}\)

4) 45%.x =\(\dfrac{3}{5}\)

=> \(x=\dfrac{3}{5}:\dfrac{9}{20}\)

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

a: \(=\dfrac{-12}{7}\left(\dfrac{4}{35}+\dfrac{31}{35}\right)-\dfrac{2}{7}=\dfrac{-12}{7}-\dfrac{2}{7}=-2\)

b: =(-4)+(-4)+...+(-4)

=-4*25=-100

c: \(=157\cdot\left(-37\right)-41\cdot53+37\cdot157+51\cdot53\)

=10*53

=530

a: \(=\dfrac{157}{8}\cdot\dfrac{12}{7}-\dfrac{61}{4}\cdot\dfrac{12}{7}\)

\(=\dfrac{12}{7}\left(\dfrac{157}{8}-\dfrac{122}{8}\right)\)

\(=\dfrac{12}{7}\cdot\dfrac{35}{8}=5\cdot\dfrac{3}{2}=\dfrac{15}{2}\)

b: \(=\dfrac{2}{15}-\dfrac{2}{15}\cdot5+\dfrac{3}{15}\)

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

c: \(=\left(\dfrac{10}{3}+\dfrac{5}{2}\right):\left(\dfrac{19}{6}-\dfrac{21}{5}\right)-\dfrac{11}{31}\)

\(=\dfrac{35}{6}:\dfrac{-31}{30}-\dfrac{11}{31}\)

\(=\dfrac{35}{6}\cdot\dfrac{30}{-31}-\dfrac{11}{31}\)

\(=\dfrac{-35\cdot5-11}{31}=\dfrac{-186}{31}=-6\)

\(A=\left(-\dfrac{43}{51}\right)\left(-\dfrac{19}{80}\right)\)

=>A>0(1)

\(B=\left(-\dfrac{7}{13}\right)\left(-\dfrac{4}{65}\right)\left(-\dfrac{8}{21}\right)\)

=>B<0(2)

C\(=-\dfrac{5}{10}.\left(-\dfrac{4}{10}\right).....\left(\dfrac{4}{10}\right)\left(\dfrac{5}{10}\right)=0\)

=>C=0(3)

Từ 1;2;3 =>A>C>B

\(A=\dfrac{-43}{51}.\dfrac{-19}{80}\Leftrightarrow A>0\left(1\right)\)

\(B=\left(\dfrac{-7}{13}\right).\left(-\dfrac{4}{65}\right).\left(\dfrac{-8}{31}\right)\Leftrightarrow B< 0\left(2\right)\)

\(C=\dfrac{-5}{10}.\dfrac{-4}{10}...........\dfrac{3}{10}.\dfrac{4}{10}.\dfrac{5}{10}\Leftrightarrow C=0\left(3\right)\)

Từ \(\left(1\right)+\left(2\right)+\left(3\right)\Leftrightarrow A>C>B\)