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a: Ta có:
\(\left(\dfrac{2}{5}+\dfrac{1}{5}\right)+\dfrac{1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
\(\dfrac{2}{5}+\left(\dfrac{1}{5}+\dfrac{1}{5}\right)=\dfrac{2}{5}+\dfrac{2}{5}=\dfrac{4}{5}\)
\(\dfrac{4}{5}=\dfrac{4}{5}\). Vậy \(\left(\dfrac{2}{5}+\dfrac{1}{5}\right)+\dfrac{1}{5}=\dfrac{2}{5}+\left(\dfrac{1}{5}+\dfrac{1}{5}\right)\)
Ta có:
\(\left(\dfrac{2}{9}+\dfrac{5}{9}\right)+\dfrac{1}{9}=\dfrac{7}{9}+\dfrac{1}{9}=\dfrac{8}{9}\)
\(\dfrac{2}{9}+\left(\dfrac{5}{9}+\dfrac{1}{9}\right)=\dfrac{2}{9}+\dfrac{6}{9}=\dfrac{8}{9}\)
\(\dfrac{8}{9}=\dfrac{8}{9}\). Vậy \(\left(\dfrac{2}{9}+\dfrac{5}{9}\right)+\dfrac{1}{9}=\dfrac{2}{9}+\left(\dfrac{5}{9}+\dfrac{1}{9}\right)\)
b: \(\left(\dfrac{1}{3}+\dfrac{2}{3}\right)+\dfrac{4}{3}=\dfrac{3}{3}+\dfrac{4}{3}=\dfrac{7}{3}\)
\(\dfrac{1}{3}+\left(\dfrac{2}{3}+\dfrac{4}{3}\right)=\dfrac{1}{3}+\dfrac{6}{3}=\dfrac{7}{3}\)
\(\dfrac{7}{3}=\dfrac{7}{3}\). Vậy \(\left(\dfrac{1}{3}+\dfrac{2}{3}\right)+\dfrac{4}{3}=\dfrac{1}{3}+\left(\dfrac{2}{3}+\dfrac{4}{3}\right)\)
Đề của anh bị sai mới đúng chứ ạ? Anh Đạt ghi là \(\left(\dfrac{2}{9}+\dfrac{5}{9}\right)+\dfrac{1}{9}\) chứ có phải \(\dfrac{2}{5}\) đâu ạ?
5 phân số đó là:
60/100
61/100
62/100
63/100
64/100
Chúc bạn học tốt nha(^-^)
a, \(=\dfrac{1+4}{5}+\dfrac{5+1+3}{9}=1+1=2\)
b, \(=\dfrac{1+4+2}{3}+\dfrac{1+2+5}{6}=\dfrac{6}{3}+\dfrac{8}{6}=2+\dfrac{4}{3}=\dfrac{6+4}{3}=\dfrac{10}{3}\)
2/3 : 1/2 + 1/3 : 1/2
= 4/3 + 2/3
= 2
3/5 : 4/7 : 7/6
= 21/20 : 7/6
= 9/10
4/9 x 7/5 : 7/9
= 28/45 : 7/9
= 4/5
Em nhớ k anh nha
có thể tính nốt phép tính 3/4 x 3/2 + 3/4 x 1/2 có đc ko ạ
\(a,\frac{2001}{2002}.\frac{5}{7}.\frac{2002}{5}.\frac{7}{2001}=\left(\frac{2001}{2002}.\frac{7}{2001}\right).\left(\frac{5}{7}.\frac{2002}{5}\right)\)
\(=\frac{7}{2002}.\frac{2002}{7}=1\)
\(b,\frac{5}{7}.\frac{7}{9}.\frac{9}{11}.\frac{11}{13}=\left(\frac{5}{7}.\frac{7}{9}\right).\left(\frac{9}{11}.\frac{11}{13}\right)=\frac{5}{9}.\frac{9}{13}\)
\(=\frac{5}{13}\)
\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(=\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)
=\(\dfrac{64}{192}+\dfrac{32}{192}+\dfrac{16}{192}+\dfrac{8}{192}+\dfrac{4}{192}+\dfrac{2}{192}+\dfrac{1}{192}\)
= \(\dfrac{127}{192}\)