a) \(1+\dfrac{4}{9}=\dfrac{9}{9}+\dfrac{4}{9}=\dfrac{9+4}{9}=\dfrac{13}{9}\)

b) \(5+\dfrac{1}{2}=\dfrac{10}{2}+\dfrac{1}{2}=\dfrac{10+1}{2}=\dfrac{11}{2}\)

c) \(3-\dfrac{5}{6}=\dfrac{18}{6}-\dfrac{5}{6}=\dfrac{18-5}{6}=\dfrac{13}{6}\)

d) \(\dfrac{31}{7}-2=\dfrac{31}{7}-\dfrac{14}{7}=\dfrac{31-14}{7}=\dfrac{17}{7}\)

a) \(\dfrac{21}{15}\) + \(\dfrac{2}{5}\) = \(\dfrac{9}{5}\)
b)  \(\dfrac{6}{16}\) + \(\dfrac{1}{8}\) = \(\dfrac{1}{2}\)

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

a: \(\dfrac{15}{8}-\dfrac{13}{8}=\dfrac{15-13}{8}=\dfrac{2}{8}=\dfrac{1}{4}\)

b: \(\dfrac{7}{15}-\dfrac{2}{15}=\dfrac{7-2}{15}=\dfrac{5}{15}=\dfrac{1}{3}\)

c: \(\dfrac{11}{12}-\dfrac{2}{12}=\dfrac{11-2}{12}=\dfrac{9}{12}=\dfrac{3}{4}\)

d: \(\dfrac{19}{7}-\dfrac{5}{7}=\dfrac{19-5}{7}=\dfrac{14}{7}=2\)

a) \(\dfrac{2}{3}=\dfrac{2.8}{3.8}=\dfrac{16}{24}\)

\(\dfrac{5}{8}=\dfrac{5.3}{8.3}=\dfrac{15}{24}\)

b) \(\dfrac{1}{4}=\dfrac{1.3}{4.3}=\dfrac{3}{12}\)

\(\dfrac{7}{12}=\dfrac{7.1}{12.1}=\dfrac{7}{12}\)

c) \(\dfrac{5}{6}=\dfrac{5.4}{6.4}=\dfrac{20}{24}\)

\(\dfrac{3}{8}=\dfrac{3.3}{8.3}=\dfrac{9}{24}\)

a) 

b) 

c) 

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Đ V T H

a: \(A=\dfrac{7}{12}+\dfrac{5}{72}-\dfrac{11}{36}=\dfrac{42}{72}+\dfrac{5}{72}-\dfrac{22}{72}=\dfrac{25}{72}\)

b: \(B=\dfrac{8+5}{10}:\dfrac{-5}{13}=\dfrac{13}{10}\cdot\dfrac{13}{-5}=-\dfrac{169}{100}\)

c: \(C=\left(\dfrac{88}{132}-\dfrac{33}{132}+\dfrac{60}{132}\right):\left(\dfrac{55}{132}+\dfrac{132}{132}-\dfrac{84}{132}\right)\)

\(=\dfrac{88-33+60}{55+132-84}=\dfrac{115}{103}\)

\(\dfrac{2}{5}+\dfrac{1}{5}=\dfrac{2+1}{5}=\dfrac{3}{5}\)

\(\dfrac{2}{3}+\dfrac{5}{3}=\dfrac{2+5}{3}=\dfrac{7}{3}\)

\(\dfrac{3}{8}+\dfrac{4}{8}=\dfrac{3+4}{8}=\dfrac{7}{8}\)

\(\dfrac{6}{9}+\dfrac{2}{9}=\dfrac{6+2}{9}=\dfrac{8}{9}\)

\(\dfrac{12}{18}+\dfrac{7}{18}=\dfrac{12+7}{18}=\dfrac{19}{18}\)

\(\dfrac{7}{4}+\dfrac{2}{4}=\dfrac{7+2}{4}=\dfrac{9}{4}\)

1: Ta có: \(\dfrac{x+4}{4}+\dfrac{3x-7}{5}=\dfrac{7x+2}{20}\)

\(\Leftrightarrow5x+20+12x-28=7x+2\)

\(\Leftrightarrow17x-7x=2+8=10\)

hay x=1

2: Ta có: \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)

\(\Leftrightarrow\dfrac{6x}{36}+\dfrac{4\left(1-3x\right)}{36}=\dfrac{3\left(-x+1\right)}{36}\)

\(\Leftrightarrow6x+4-12x=-3x+3\)

\(\Leftrightarrow-6x+3x=3-4\)

hay \(x=\dfrac{1}{3}\)

3: Ta có: \(\dfrac{x-3}{3}-\dfrac{x+2}{12}=\dfrac{2x-1}{4}\)

\(\Leftrightarrow4x-12-x-2=6x-3\)

\(\Leftrightarrow3x-14-6x+3=0\)

\(\Leftrightarrow-3x=11\)

hay \(x=-\dfrac{11}{3}\)

4: Ta có: \(\dfrac{x-2}{4}-\dfrac{2x+3}{3}=\dfrac{x+6}{12}\)

\(\Leftrightarrow3x-6-8x-12=x+6\)

\(\Leftrightarrow-5x-x=6+18\)

hay x=-4

5: Ta có: \(\dfrac{2x-1}{12}-\dfrac{3-x}{18}=\dfrac{-1}{36}\)

\(\Leftrightarrow6x-3+2x-6=-1\)

\(\Leftrightarrow8x=8\)

hay x=1