Tính (theo mẫu):
a) \(\dfrac{1}{4}\) của 20 km. b) \(\dfrac{1}{7}\) của 28 g.
c) \(\dfrac{3}{10}\) của 100 ml. d) \(\dfrac{3}{4}\) của 640 tấn.
e) \(\dfrac{5}{8}\) của 40 m2. g) \(\dfrac{2}{3}\) của 1 giờ.
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 \(\dfrac{1}{8}:\dfrac{1}{5}=\dfrac{5}{8}\)
đúng ghi Đ,sai ghi S
a)\(\dfrac{2}{3}\) của một nửa là \(\dfrac{1}{3}\)
b)\(\dfrac{1}{5}\) của \(\dfrac{1}{4}\) là \(\dfrac{1}{20}\)
c)Một nửa của \(\dfrac{1}{2}\) là \(\dfrac{1}{4}\)
d)\(\dfrac{2}{5}\) của \(\dfrac{4}{7}\) là \(\dfrac{7}{10}\)
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: \(\dfrac{23}{100}+\dfrac{27}{100}=\dfrac{23+27}{100}=\dfrac{50}{100}=\dfrac{1}{2}\)
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: \(\dfrac{19}{7}-\dfrac{5}{7}=\dfrac{19-5}{7}=\dfrac{14}{7}=2\)
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
Quy đồng mẫu các phân số sau:
a) \(\dfrac{3}{4}\), \(\dfrac{1}{-9}\),\(\dfrac{-5}{5}\) b)\(\dfrac{1}{-7}\), \(\dfrac{-1}{-8}\), \(\dfrac{3}{4}\)
c)\(\dfrac{1}{3}\),\(\dfrac{-1}{5}\), \(\dfrac{1}{-10}\) d)\(\dfrac{5}{14}\),\(\dfrac{-3}{7}\),\(\dfrac{3}{-4}\)
e)\(\dfrac{4}{11}\),\(\dfrac{3}{-5}\),\(\dfrac{-1}{-2}\) g)\(\dfrac{7}{15}\),\(\dfrac{-3}{8}\),\(\dfrac{-2}{3}\)
a: 3/4=135/180
-1/9=-20/180
-5/5=-1=-180/180
Bài 1: Tính giá trị của biểu thức sau
A=1-\(\dfrac{50-\dfrac{4}{2018}+\dfrac{2}{2019}-\dfrac{2}{2020}}{100-\dfrac{8}{2018} +\dfrac{4}{2019}-\dfrac{4}{2020}}\)
B=\(\dfrac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
C=\(x^{2020}\)-\(y^{2020}\)+\(xy^{2019}\)-\(x^{2019}\).y+2019 biết x-y=0
Mong mn giúp đỡ
a: \(A=1-\dfrac{2\left(25-\dfrac{2}{2018}+\dfrac{1}{2019}-\dfrac{1}{2020}\right)}{4\left(25-\dfrac{2}{2018}+\dfrac{1}{2019}-\dfrac{1}{2020}\right)}\)
=1-2/4=1/2
b: \(B=\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}\)
\(=\dfrac{5^{10}\cdot7^3\left(1-7\right)}{5^9\cdot7^3\left(1+2^3\right)}=5\cdot\dfrac{-6}{9}=-\dfrac{10}{3}\)
c: x-y=0 nên x=y
\(C=x^{2020}-x^{2020}+y\cdot y^{2019}-y^{2019}\cdot y+2019\)
=2019
3. Tính :
a/ \(\dfrac{-1}{2}\) + \(\dfrac{5}{6}\) + \(\dfrac{1}{3}\) b/ \(\dfrac{-3}{8}\) + \(\dfrac{7}{4}\) - \(\dfrac{1}{12}\) c/ \(\dfrac{3}{5}\) : (\(\dfrac{1}{4}\) . \(\dfrac{7}{5}\)) d/ \(\dfrac{10}{11}\) + \(\dfrac{4}{11}\) : 4 - \(\dfrac{1}{8}\)
3.a)\(\dfrac{-1}{2}+\dfrac{5}{6}+\dfrac{1}{3}=\dfrac{-3}{6}+\dfrac{5}{6}+\dfrac{2}{6}=\dfrac{-3+5+2}{6}=\dfrac{4}{6}=\dfrac{2}{3}\)
b)\(\dfrac{-3}{8}+\dfrac{7}{4}-\dfrac{1}{12}=\dfrac{-9}{24}+\dfrac{42}{24}-\dfrac{2}{24}=\dfrac{-9+42-2}{24}=\dfrac{31}{24}\)
c)\(\dfrac{3}{5}:\left(\dfrac{1}{4}.\dfrac{7}{5}\right)=\dfrac{3}{5}:\dfrac{7}{20}=\dfrac{3}{5}.\dfrac{20}{7}=\dfrac{12}{7}\)
d)\(\dfrac{10}{11}+\dfrac{4}{11}:4-\dfrac{1}{8}=\dfrac{10}{11}+\dfrac{4}{11}.\dfrac{1}{4}-\dfrac{1}{8}=\dfrac{10}{11}+\dfrac{1}{11}-\dfrac{1}{8}=1-\dfrac{1}{8}=\dfrac{8}{8}-\dfrac{1}{8}=\dfrac{7}{8}\)