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) \(\dfrac{2}{5}-\dfrac{3}{20}=\dfrac{8}{20}-\dfrac{3}{20}=\dfrac{8-3}{20}=\dfrac{5}{20}=\dfrac{1}{4}\)
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: \(\dfrac{23}{100}+\dfrac{27}{100}=\dfrac{23+27}{100}=\dfrac{50}{100}=\dfrac{1}{2}\)
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
Tính (theo mẫu).
Mẫu: \(\dfrac{2}{5}\times3=\dfrac{2}{5}\times\dfrac{3}{1}=\dfrac{2\times3}{5\times1}=\dfrac{6}{5}\) Ta có thể viết gọn như sau: \(\dfrac{2}{5}\times3=\dfrac{2\times3}{5}=\dfrac{6}{5}\) |
a) \(\dfrac{9}{11}\times8\) b) \(\dfrac{4}{5}\times1\) c) \(\dfrac{15}{8}\times0\)
a) \(\dfrac{9}{11}\times8=\dfrac{9\times8}{11}=\dfrac{72}{11}\)
b) \(\dfrac{4}{5}\times1=\dfrac{4\times1}{5}=\dfrac{4}{5}\)
c) \(\dfrac{15}{8}\times0=\dfrac{15\times0}{8}=\dfrac{0}{8}=0\)
a: 9/11*8=(9*8)/11=72/11
b: 4/5*1=(4*1)/5=4/5
c: 15/8*0=(15*0)/8=0/8=0
Quy đồng mẫu số :
\(\dfrac{1}{2}+\dfrac{2}{4}\)
\(\dfrac{3}{-12}+\dfrac{4}{6}\)
\(\dfrac{1}{2}+\dfrac{2}{4}\\ \dfrac{2}{4}=\dfrac{1\cdot2}{2\cdot2}=\dfrac{2}{4}\\ \dfrac{3}{-12}+\dfrac{4}{12}\\ \dfrac{3}{-12}=\dfrac{-3}{12}\\ \dfrac{4}{6}=\dfrac{4\cdot2}{6\cdot2}=\dfrac{8}{12}\\ \dfrac{-3}{12}+\dfrac{8}{13}=\dfrac{5}{12}\)
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 cách thuận tiện :
\(\dfrac{13}{7}+\dfrac{5}{6}+\dfrac{2}{7}+\dfrac{7}{6}\) = ?
\(\dfrac{1}{2}x\dfrac{5}{6}+\dfrac{1}{2}x\dfrac{11}{6}=?\)
\(\dfrac{13}{7}+\dfrac{5}{6}+\dfrac{2}{7}+\dfrac{7}{6}=\dfrac{15}{7}+\dfrac{12}{6}=\dfrac{29}{7}\)
\(\dfrac{1}{2}\times\dfrac{5}{6}+\dfrac{1}{2}\times\dfrac{11}{6}=\dfrac{1}{2}\times\left(\dfrac{5}{6}+\dfrac{11}{6}\right)=\dfrac{1}{2}\times\dfrac{16}{6}=\dfrac{4}{3}\)
Thực hiện phép tính, tính nhanh nếu có thể
a, \(\dfrac{7}{13}\) + \(\dfrac{12}{13}\) + \(\dfrac{6}{-13}\)
b, ( \(\dfrac{4}{5}\) + \(\dfrac{1}{2}\) ) x ( \(\dfrac{6}{13}\) - 2 )
c, 75% : \(\dfrac{1}{2}\) - \(2\dfrac{1}{2}\) x ( \(-1\dfrac{1}{3}\) )
d, 0,375 x \(1\dfrac{3}{5}\) + 60% x \(\dfrac{2}{7}\) + \(\dfrac{3}{5}\) x \(\dfrac{5}{7}\)
e, \(\dfrac{-5}{6}\) x \(\dfrac{7}{13}\) + \(\dfrac{19}{13}\) : \(\dfrac{6}{-5}\) + \(\dfrac{2}{5}\)
f, 1,25 x \(\dfrac{7}{9}+\dfrac{5}{4}x\dfrac{15}{9}-1\dfrac{1}{4}:\dfrac{19}{3}\)
a: \(=\dfrac{7+12-6}{13}=1\)
b: \(=\dfrac{13}{10}\cdot\dfrac{6-26}{13}=\dfrac{-20}{10}=-2\)
c: \(=\dfrac{3}{4}\cdot2-\dfrac{5}{2}\cdot\dfrac{-4}{3}=\dfrac{3}{2}+\dfrac{20}{6}=\dfrac{3}{2}+\dfrac{10}{3}=\dfrac{29}{6}\)
d: \(=\dfrac{3}{8}\cdot\dfrac{8}{5}+\dfrac{3}{5}\cdot\dfrac{2}{7}+\dfrac{3}{5}\cdot\dfrac{5}{7}=\dfrac{3}{5}+\dfrac{3}{5}=\dfrac{6}{5}\)
Bài 4. Trên tia Ox lấy điểm A và B sao cho OA = 8cm, OB = 12cm.
a, Điểm A có nằm giữa hai điểm O và B không? Vì sao? Tính AB.
b, Gọi M, N lần lượt là trung điểm của OA, OB. Điểm M có nằm giữa hai điểm O và N không? Vì sao? Tính MN.
c, Điểm N có là trung điểm của đoạn thẳng AM không? Vì sao?
cho mìn hỏi câu b nhoa
a) Cho phân số \(\dfrac{13}{42}\). Hãy tìm một số tự nhiên n sao cho khi cộng tử số với n và giữ nguyên mẫu số thì được phân số mới có giá trị bằng \(\dfrac{5}{6}\).
b) Tính nhanh
\(\dfrac{1}{2}+\dfrac{2}{4}+\dfrac{3}{6}+\dfrac{4}{8}+\dfrac{5}{10}+\dfrac{6}{12}+\dfrac{7}{14}+\dfrac{8}{16}+\dfrac{9}{18}+\dfrac{10}{20}\)
