C

C

c

\(y=12-\left(\dfrac{1}{3}\times4\right)=12-\dfrac{4}{3}=\dfrac{32}{3}\)

\(y=\left(\dfrac{12}{35}\times\dfrac{14}{9}\right)-\dfrac{2}{5}=\dfrac{8}{15}-\dfrac{2}{5}=\dfrac{2}{15}\)

a) \(y+\dfrac{1}{3}\times4=12\)

\(y+\dfrac{1}{3}=12:4\)

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

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

\(y=\dfrac{8}{3}\)

b) \(\dfrac{2}{5}+y=\dfrac{12}{35}\times\dfrac{14}{9}\)

    \(\dfrac{2}{5}+y=\dfrac{8}{15}\)

           \(y=\dfrac{8}{15}-\dfrac{2}{5}\)

          \(y=\dfrac{2}{15}\)

2/15

\(-1\dfrac{1}{5}.\dfrac{12+\dfrac{4}{3}-\dfrac{12}{37}-\dfrac{12}{35}}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{35}}:\dfrac{4+\dfrac{4}{17}+\dfrac{4}{19}+\dfrac{4}{2003}}{5+\dfrac{5}{17}+\dfrac{5}{19}+\dfrac{5}{2003}}\)

\(=\dfrac{-6}{5}.\dfrac{4\left(3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{35}\right)}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{35}}:\dfrac{4\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2003}\right)}{5\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2003}\right)}\)

\(=\dfrac{-6}{5}.4:\dfrac{4}{5}\)

\(=\dfrac{-6.4.5}{5.4}=-6\)

= -4 đúng không

\(-1\dfrac{1}{5}.\dfrac{12+\dfrac{4}{3}-\dfrac{12}{37}-\dfrac{12}{35}}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}}:\dfrac{4+\dfrac{4}{17}+\dfrac{4}{19}+\dfrac{4}{2003}}{5+\dfrac{5}{17}+\dfrac{5}{19}+\dfrac{5}{2003}}\)

\(=\dfrac{-6}{5}.\dfrac{4\left(3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}\right)}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}}:\dfrac{4\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2003}\right)}{5\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2003}\right)}\)

\(=\dfrac{-6}{5}.\dfrac{4}{3}:\dfrac{4}{5}\)

\(=\dfrac{-6.4.5}{5.3.4}=\dfrac{-6}{3}=-2\)

Vậy...

a: Ta có: \(\dfrac{8}{9}-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{72}\right)\)

\(=\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{3}-...+\dfrac{1}{8}-\dfrac{1}{9}\right)\)

=0

1. \(-\dfrac{52}{17}+\left(\dfrac{12}{19}+\dfrac{52}{17}\right)=\left[\left(-\dfrac{52}{17}\right)+\left(-\dfrac{52}{17}\right)\right]+\dfrac{12}{19}=\dfrac{12}{19}\)

2. \(\dfrac{21}{35}+\left(-1+\dfrac{14}{35}\right)=\dfrac{3}{5}+\left(-1+\dfrac{2}{5}\right)=\left(\dfrac{3}{5}+\dfrac{2}{5}\right)+\left(-1\right)=1-1=0\)

1. \(\frac{-52}{17}+\left(\frac{12}{19}+\frac{52}{17}\right)=\frac{-52}{17}+\frac{12}{19}+\frac{52}{17}\)\(=\left(\frac{-52}{17}+\frac{52}{17}\right)+\frac{12}{19}=\frac{0}{17}+\frac{12}{19}=\frac{12}{19}\)

2. \(\frac{21}{35}+\left(-1+\frac{14}{35}\right)=\frac{21}{35}-1+\frac{14}{35}=\left(\frac{21}{35}+\frac{14}{35}\right)-1=1-1=0\)

a/ \(x-\dfrac{3}{7}=\dfrac{2}{5}\cdot\dfrac{1}{4}\)

\(x-\dfrac{3}{7}=\dfrac{1}{10}\)

\(x=\dfrac{1}{10}+\dfrac{3}{7}=\dfrac{37}{70}\)

Vậy....

b/ \(x+\dfrac{4}{5}=-\dfrac{5}{12}\cdot\dfrac{3}{25}\)

\(x+\dfrac{4}{5}=-\dfrac{1}{20}\)

\(x=-\dfrac{1}{20}-\dfrac{4}{5}=-\dfrac{17}{20}\)

Vậy....

c/ \(\dfrac{x}{182}=-\dfrac{6}{12}\cdot\dfrac{35}{91}\)

\(\dfrac{x}{182}=-\dfrac{5}{26}\)

\(=>x\cdot26=-5\cdot182\)

\(26x=-910\)

\(x=-910:26=-35\)

Vậy....

a) Ta có: \(x-\dfrac{3}{7}=\dfrac{2}{5}\cdot\dfrac{1}{4}\)

\(\Leftrightarrow x-\dfrac{3}{7}=\dfrac{1}{10}\)

\(\Leftrightarrow x=\dfrac{1}{10}+\dfrac{3}{7}=\dfrac{7}{70}+\dfrac{30}{70}\)

hay \(x=\dfrac{37}{70}\)

Vậy: \(x=\dfrac{37}{70}\)

a. \(\dfrac{-3}{12}+\dfrac{1}{-4}=\dfrac{-3}{12}+\dfrac{-3}{12}=\dfrac{-3-3}{12}=\dfrac{-6}{12}=\dfrac{-1}{2}\)
 

b. \(\dfrac{5}{12}+\dfrac{-3}{28}=\dfrac{35}{84}+\dfrac{-9}{84}=\dfrac{35+\left(-9\right)}{84}=\dfrac{26}{84}=\dfrac{13}{42}\)
 

c. \(\dfrac{-7}{15}+\dfrac{3}{35}=\dfrac{-49}{105}+\dfrac{9}{105}=\dfrac{-49+9}{105}=\dfrac{-40}{105}=\dfrac{-8}{21}\)
 

d. \(\dfrac{-5}{7}+\dfrac{-3}{4}=\dfrac{-20}{28}+\dfrac{-21}{28}=\dfrac{-20+\left(-21\right)}{28}=\dfrac{-41}{28}\)

\(\dfrac{11}{16}\) = \(\dfrac{1}{16}\) + \(\dfrac{10}{16}\) = \(\dfrac{1}{16}\) + \(\dfrac{2+8}{2\times8}\) = \(\dfrac{1}{16}\) + \(\dfrac{2}{2\times8}\) + \(\dfrac{8}{2\times8}\) = \(\dfrac{1}{16}\) + \(\dfrac{1}{8}\)+ \(\dfrac{1}{2}\)

\(\dfrac{5}{12}\) = \(\dfrac{1}{12}\) + \(\dfrac{4}{12}\) = \(\dfrac{1}{12}\) + \(\dfrac{4}{3\times4}\) = \(\dfrac{1}{12}\) + \(\dfrac{1}{3}\)

\(\dfrac{6}{35}\) = \(\dfrac{1}{35}\) + \(\dfrac{5}{35}\) = \(\dfrac{1}{35}\) + \(\dfrac{5}{5\times7}\) = \(\dfrac{1}{35}\) + \(\dfrac{1}{7}\)