Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{7}{18}\)+ \(\frac{5}{3}\)* 1\(\frac{3}{9}\): \(\frac{55}{4}\)
Mọi người giúp mình giải nhé!!!
\(\frac{7}{18}+\frac{5}{3}\times1\frac{3}{9}\div\frac{55}{4}=\)\(\frac{7}{18}+\frac{5}{3}\times\frac{12}{9}\div\frac{55}{4}=\)\(\frac{7}{18}+\frac{5}{3}\times\frac{12}{9}\times\frac{4}{55}=\)\(\frac{7}{18}+\frac{5\times12\times4}{3\times9\times55}=\)\(\frac{7}{18}+\frac{4\times4}{9\times11}=\)\(\frac{7}{18}+\frac{16}{99}=\)\(\frac{693}{1782}+\frac{288}{1782}=\)\(\frac{981}{1782}=\frac{109}{198}\)
Câu hỏi của Mun - Toán lớp 5 - Học toán với OnlineMath
Bạn vào đây xem
A = 2011.2010−12009.2011+2010
=> \(\frac{2011.2009+2011-1}{2009.2011+2010}\) = \(\frac{2011.2009+2010}{2009.2011+2010}=1\)
=> A = 1
t i c k nhé!! 354543564576856925253456456545756756887676976
a,(11/15+4/15)+(5/7+2/7)
=1+1
=2
b,5/9x(1/2+6/4)
=5/9x2
=10/9
c,1/2:(7/8+9/8)
=1/2:2
=1
d,(17/10-7/10)+1/2
=1+1/2
=3/2
a) \(\frac{11}{15}+\frac{5}{7}+\frac{2}{7}+\frac{4}{15}=\left(\frac{11}{15}+\frac{4}{15}\right)+\left(\frac{5}{7}+\frac{2}{7}\right)\)
\(=2\)
b) \(\frac{5}{9}\times\frac{1}{2}\times\frac{5}{9}\times\frac{6}{4}=\frac{25}{81}\times\frac{3}{4}=\frac{25}{108}\)
c) \(\frac{7}{8}\div\frac{1}{2}+\frac{9}{8}\div\frac{1}{2}=\left(\frac{7}{8}+\frac{9}{8}\right)\div\frac{1}{2}\)
\(=2\div\frac{1}{2}=4\)
d) \(\frac{17}{10}+\frac{1}{2}-\frac{7}{10}=\left(\frac{17}{10}-\frac{7}{10}\right)+\frac{1}{2}\)
\(=1+\frac{1}{2}=\frac{3}{2}\)
a) \(\frac{11}{15}+\frac{5}{7}+\frac{2}{7}+\frac{4}{15}\)
\(=\left(\frac{11}{15}+\frac{4}{15}\right)+\left(\frac{5}{7}+\frac{2}{7}\right)\)
\(=1+1\)
\(=2\)
b) \(\frac{5}{9}.\frac{1}{2}.\frac{5}{9}.\frac{6}{4}\)
\(=\left(\frac{5}{9}\right)^2\left(\frac{1}{2}.\frac{6}{4}\right)\)
\(=\frac{25}{81}.\frac{3}{4}\)
\(=\frac{25}{108}\)
c) \(\frac{7}{8}:\frac{1}{2}+\frac{9}{8}:\frac{1}{2}\)
\(=\frac{7}{8}.2+\frac{9}{8}.2\)
\(=2\left(\frac{7}{8}+\frac{9}{8}\right)\)
\(=2.\frac{16}{8}\)
\(=2.2\)
\(=4\)
d) \(\frac{17}{10}+\frac{1}{2}-\frac{7}{10}\)
\(=\left(\frac{17}{10}-\frac{7}{10}\right)+\frac{1}{2}\)
\(=1+\frac{1}{2}\)
\(=\frac{2}{2}+\frac{1}{2}\)
\(=\frac{3}{2}\)