Tính (theo mẫu).
a)
\(\dfrac{3}{5}+\dfrac{1}{10}=?\)
b)
\(\dfrac{8}{9}+\dfrac{2}{3}=?\)
c)
\(\dfrac{1}{2}+\dfrac{5}{8}=?\)
Tính rồi rút gọn (theo mẫu):
Mẫu: \(\dfrac{5}{6}+\dfrac{4}{6}=\dfrac{5+4}{6}=\dfrac{9}{6}=\dfrac{3}{2}\) |
a) \(\dfrac{1}{8}+\dfrac{5}{8}\) b) \(\dfrac{1}{15}+\dfrac{4}{15}\) c) \(\dfrac{5}{9}+\dfrac{7}{9}\) d) \(\dfrac{23}{100}+\dfrac{27}{100}\)
a: \(\dfrac{1}{8}+\dfrac{5}{8}=\dfrac{1+5}{8}=\dfrac{6}{8}=\dfrac{3}{4}\)
b: \(\dfrac{1}{15}+\dfrac{4}{15}=\dfrac{1+4}{15}=\dfrac{5}{15}=\dfrac{1}{3}\)
c: \(\dfrac{5}{9}+\dfrac{7}{9}=\dfrac{5+7}{9}=\dfrac{12}{9}=\dfrac{4}{3}\)
d: \(\dfrac{23}{100}+\dfrac{27}{100}=\dfrac{23+27}{100}=\dfrac{50}{100}=\dfrac{1}{2}\)
Tính rồi rút gọn (theo mẫu):
Mẫu: \(\dfrac{9}{10}-\dfrac{4}{10}=\dfrac{9-4}{10}=\dfrac{5}{10}=\dfrac{1}{2}\) |
a) \(\dfrac{15}{8}-\dfrac{13}{8}\) b) \(\dfrac{7}{15}-\dfrac{2}{15}\) c) \(\dfrac{11}{12}-\dfrac{2}{12}\) d) \(\dfrac{19}{7}-\dfrac{5}{7}\)
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: \(\dfrac{19}{7}-\dfrac{5}{7}=\dfrac{19-5}{7}=\dfrac{14}{7}=2\)
Tính (theo mẫu):
a)
\(5+\dfrac{3}{2}\) \(\dfrac{3}{4}+2\) \(\dfrac{8}{9}+3\)
b)
\(1-\dfrac{1}{2}\) \(5-\dfrac{7}{3}\) \(\dfrac{11}{2}-3\)
a: \(5+\dfrac{3}{2}=\dfrac{10}{2}+\dfrac{3}{2}=\dfrac{10+3}{2}=\dfrac{13}{2}\)
\(\dfrac{3}{4}+2=\dfrac{3}{4}+\dfrac{8}{4}=\dfrac{3+8}{4}=\dfrac{11}{4}\)
\(\dfrac{8}{9}+3=\dfrac{8}{9}+\dfrac{27}{9}=\dfrac{8+27}{9}=\dfrac{35}{9}\)
b: \(1-\dfrac{1}{2}=\dfrac{2}{2}-\dfrac{1}{2}=\dfrac{2-1}{2}=\dfrac{1}{2}\)
\(5-\dfrac{7}{3}=\dfrac{15}{3}-\dfrac{7}{3}=\dfrac{15-7}{3}=\dfrac{8}{3}\)
\(\dfrac{11}{2}-3=\dfrac{11}{2}-\dfrac{6}{2}=\dfrac{11-6}{2}=\dfrac{5}{2}\)
Tính (theo mẫu).
Mẫu: \(\dfrac{1}{2}-\dfrac{5}{12}=\dfrac{6}{12}-\dfrac{5}{12}=\dfrac{6-5}{12}=\dfrac{1}{12}\) |
a) \(\dfrac{3}{4}-\dfrac{1}{8}\) b) \(\dfrac{2}{6}-\dfrac{5}{18}\) c) \(\dfrac{2}{5}-\dfrac{3}{20}\)
a) \(\dfrac{3}{4}-\dfrac{1}{8}=\dfrac{6}{8}-\dfrac{1}{8}=\dfrac{6-1}{8}=\dfrac{5}{8}\)
b) \(\dfrac{2}{6}-\dfrac{5}{18}=\dfrac{6}{18}-\dfrac{5}{18}=\dfrac{6-5}{18}=\dfrac{1}{18}\)
c) \(\dfrac{2}{5}-\dfrac{3}{20}=\dfrac{8}{20}-\dfrac{3}{20}=\dfrac{8-3}{20}=\dfrac{5}{20}=\dfrac{1}{4}\)
Rút gọn rồi tính (theo mẫu).
Mẫu: \(\dfrac{5}{15}+\dfrac{4}{3}=\dfrac{1}{3}+\dfrac{4}{3}=\dfrac{1+4}{3}=\dfrac{5}{3}\) |
a) \(\dfrac{21}{15}+\dfrac{2}{5}\) b) \(\dfrac{6}{16}+\dfrac{1}{8}\) c) \(\dfrac{3}{12}+\dfrac{3}{4}\)
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
Tìm phân số thích hợp (theo mẫu).
Mẫu: \(\dfrac{3}{5}\times?=\dfrac{4}{7}\) \(\dfrac{4}{7}:\dfrac{3}{5}=\dfrac{20}{21}\) |
a) \(\dfrac{2}{5}\times?=\dfrac{3}{10}\) b) \(\dfrac{1}{8}:?=\dfrac{1}{5}\)
a) \(\dfrac{2}{5}\times?=\dfrac{3}{10}\)
\(?=\dfrac{3}{10}:\dfrac{2}{5}=\dfrac{3}{4}\)
b) \(\dfrac{1}{8}:?=\dfrac{1}{5}\)
\(?=\dfrac{1}{8}:\dfrac{1}{5}=\dfrac{5}{8}\)
a: Phân số cần tìm là: \(\dfrac{3}{10}:\dfrac{2}{5}=\dfrac{3}{10}\cdot\dfrac{5}{2}=\dfrac{15}{20}=\dfrac{3}{4}\)
b: Phân số cần tìm là \(\dfrac{1}{8}:\dfrac{1}{5}=\dfrac{5}{8}\)
Tính (theo mẫu).
Mẫu: \(5\times\dfrac{2}{9}=\dfrac{5}{1}\times\dfrac{2}{9}=\dfrac{5\times2}{1\times9}=\dfrac{10}{9}\) Ta có thể viết gọn như sau: \(5\times\dfrac{2}{9}=\dfrac{5\times2}{9}=\dfrac{10}{9}\) |
a) \(3\times\dfrac{4}{11}\) b) \(1\times\dfrac{5}{4}\) c) \(0\times\dfrac{2}{5}\)
a) \(3\times\dfrac{4}{11}=\dfrac{3\times4}{11}=\dfrac{12}{11}\)
b) \(1\times\dfrac{5}{4}=\dfrac{1\times5}{4}=\dfrac{5}{4}\)
c) \(0\times\dfrac{2}{5}=\dfrac{0\times2}{5}=\dfrac{0}{5}=0\)
a: \(=\dfrac{3\cdot4}{11}=\dfrac{12}{11}\)
b: \(=\dfrac{1\cdot5}{4}=\dfrac{5}{4}\)
c: \(=\dfrac{0\cdot2}{5}=0\)
Tính (theo mẫu).
Mẫu: \(\dfrac{2}{3}+\dfrac{5}{3}=\dfrac{2+5}{3}=\dfrac{7}{3}\)
a) \(\dfrac{2}{7}+\dfrac{4}{7}\) b) \(\dfrac{23}{13}+\dfrac{8}{13}\) c) \(\dfrac{27}{125}+\dfrac{16}{125}\)
a) \(\dfrac{2}{7}+\dfrac{4}{7}=\dfrac{2+4}{7}=\dfrac{6}{7}\)
b) \(\dfrac{23}{13}+\dfrac{8}{13}=\dfrac{23+8}{13}=\dfrac{31}{13}\)
c) \(\dfrac{27}{125}+\dfrac{16}{125}=\dfrac{27+16}{125}=\dfrac{43}{125}\)
a)\(\dfrac{2}{7}\) + \(\dfrac{4}{7}\) = \(\dfrac{6}{7}\)
b)\(\dfrac{23}{13}\) + \(\dfrac{8}{13}\) = \(\dfrac{31}{13}\)
c)\(\dfrac{27}{125}\) + \(\dfrac{16}{125}\) = \(\dfrac{43}{125}\)
Tính (theo mẫu).
Mẫu: \(2+\dfrac{1}{6}=\dfrac{12}{6}+\dfrac{1}{6}=\dfrac{13}{6};1-\dfrac{1}{4}=\dfrac{4}{4}-\dfrac{1}{4}=\dfrac{3}{4}\) |
a) \(1+\dfrac{4}{9}\) b) \(5+\dfrac{1}{2}\) c) \(3-\dfrac{5}{6}\) d) \(\dfrac{31}{7}-2\)
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) \(\dfrac{31}{7}-2=\dfrac{31}{7}-\dfrac{14}{7}=\dfrac{31-14}{7}=\dfrac{17}{7}\)