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Giải SBT Toán 5: Ôn tập: Phép cộng và phép trừ hai phân số

Bài 25 trang 7 Sách bài tập Toán lớp 5 Tập 1:

a) $\frac{3}{4}+\frac{2}{3};\frac{3}{5}+\frac{7}{10};\frac{1}{4}+\frac{7}{6}$; $\frac{2}{3}-\frac{2}{7};\frac{3}{4}-\frac{5}{12};\frac{5}{6}-\frac{2}{9}$      

b)  $2+\frac{5}{7};\frac{13}{5}-2;3-\frac{3}{8}$

c)$\frac{1}{2}+\frac{1}{3}+\frac{1}{6};\frac{5}{12}+\frac{5}{6}-\frac{3}{4};1-\left(\frac{1}{5}+\frac{1}{2}\right)$  

Lời giải:

a)  $\frac{3}{4}+\frac{2}{3}=\frac{9}{12}+\frac{8}{12}=\frac{17}{12};$

$\frac{3}{5}+\frac{7}{10}=\frac{6}{10}+\frac{7}{10}=\frac{13}{10}$

$\frac{1}{4}+\frac{7}{6}=\frac{3}{12}+\frac{14}{12}=\frac{17}{12};$

$\frac{2}{3}-\frac{2}{7}=\frac{14}{21}-\frac{6}{21}=\frac{8}{21}$                                     

 $\frac{3}{4}-\frac{5}{12}=\frac{9}{12}-\frac{5}{12}=\frac{4}{12}=\frac{1}{3};$                                         

$\frac{5}{6}-\frac{2}{9}=\frac{15}{18}-\frac{4}{18}=\frac{11}{18}$                                  

b) $2+\frac{5}{7}=\frac{14}{7}+\frac{5}{7}=\frac{19}{7}$                                          

$\frac{13}{5}-2=\frac{13}{5}-\frac{10}{5}=\frac{3}{5}$

$3-\frac{3}{8}=\frac{24}{8}-\frac{3}{8}=\frac{21}{8}$         

c)  $\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{3}{6}+\frac{2}{6}+\frac{1}{6}=\frac{6}{6}=1$

$\frac{5}{12}+\frac{5}{6}-\frac{3}{4}$

$=\frac{5}{12}+\frac{10}{12}-\frac{3}{4}$

$=\frac{15}{12}-\frac{9}{12}=\frac{6}{12}=\frac{1}{2}$    

$1-\left(\frac{1}{5}+\frac{1}{2}\right)=1-\left(\frac{2}{10}+\frac{5}{10}\right)$

$=1-\frac{7}{10}=\frac{10}{10}-\frac{7}{10}=\frac{3}{10}$