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a) \(\dfrac{2}{3}\times\dfrac{4}{5}=\dfrac{4}{5}\times\dfrac{2}{3}\)
b) \(\left(\dfrac{1}{3}\times\dfrac{2}{5}\right)\times\dfrac{3}{4}=\dfrac{1}{3}\times\left(\dfrac{2}{5}\times\dfrac{3}{4}\right)\)
c) \(\left(\dfrac{1}{3}-\dfrac{2}{15}\right)\times\dfrac{3}{4}=\dfrac{1}{3}\times\dfrac{3}{4}+\dfrac{2}{15}\times\dfrac{3}{4}\)
1 \(A=\left(1+\frac{1}{2}\right)\times\left(1+\frac{1}{3}\right)\times\left(1+\frac{1}{4}\right)\times.........\times\left(1+\frac{1}{2016}\right)\times\left(1+\frac{1}{2017}\right)\)
\(A=\frac{3}{2}\times\frac{4}{3}\times\frac{5}{4}\times......\times\frac{2016}{2017}\times\frac{2018}{2017}\)
\(A=\frac{2018}{2}=1009\)
\(B=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+.......+\frac{2}{43.45}\)
\(B=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-......+\frac{1}{43}-\frac{1}{45}\)
\(B=\frac{1}{3}-\frac{1}{45}\)
\(B=\frac{14}{45}\)
2 \(\frac{2017}{2018}\times\frac{23}{47}+\frac{24}{2018}\times\frac{2017}{47}\)
\(=\frac{2017}{2018}\times\frac{23}{47}+\frac{24}{47}\times\frac{2017}{2018}\)
\(=\frac{2017}{2018}\times\left(\frac{23}{47}+\frac{24}{47}\right)\)
\(=\frac{2017}{2018}\times1\)
=\(\frac{2017}{2018}\)
bạn nào xem giải thế có đúng ko
= 3/2 + 4/3 + 5/4 ................................ 100/99
= 100/2 = 50
a) \(\dfrac{3}{22}\times\dfrac{3}{11}\times22\)
\(=\dfrac{3}{22}\times\dfrac{3}{11}\times\dfrac{22}{1}\)
\(=\dfrac{3\times3\times22}{22\times11\times1}\)
\(=\dfrac{9}{11}\)
b) \(\left(\dfrac{1}{3}+\dfrac{1}{6}\right)\times\dfrac{2}{5}\)
\(=\left(\dfrac{2}{6}+\dfrac{1}{6}\right)\times\dfrac{2}{5}\)
\(=\dfrac{3}{6}\times\dfrac{2}{5}\)
\(=\dfrac{1}{2}\times\dfrac{2}{5}\)
\(=\dfrac{2}{10}\)
\(=\dfrac{1}{5}\)
a: =3/22*22*3/11
=3*3/11
=9/11
b: =(2/6+1/6)*2/5
=1/2*2/5
=1/5
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?
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{100}\right).200x=4036\)
\(\Leftrightarrow\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{99}{100}.200x=4036\)
\(\Leftrightarrow\frac{1.2.3...99}{2.3.4....100}.200x=4036\)
\(\Leftrightarrow\frac{1}{100}.200x=4036\)
\(\Leftrightarrow\frac{1}{100}.200x=4036\)
\(\Leftrightarrow2x=4036\)
\(\Leftrightarrow x=4036:2=2018\)
\(\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times...\times\left(1-\frac{1}{100}\right)\times200\times x=4036\)
=> \(\frac{1}{2}\times\frac{2}{3}\times...\times\frac{99}{100}\times200\times x=4036\)
=> \(\frac{1\times2\times...\times99}{2\times3\times...\times100}\times200\times x=4036\)
\(\Rightarrow\frac{1}{100}\times200\times x=4036\)
\(\Rightarrow2\times x=4036\)
=> x = 2018
a) $\frac{5}{2} \times \frac{4}{3} + \frac{1}{3} = \frac{{10}}{3} + \frac{1}{3} = \frac{{11}}{3}$
b) $\frac{7}{3} - \frac{2}{3}:\frac{5}{7} = \frac{7}{3} - \frac{2}{3} \times \frac{7}{5} = \frac{7}{3} - \frac{{14}}{{15}} = \frac{{35}}{{15}} - \frac{{14}}{{15}} = \frac{{21}}{{15}} = \frac{7}{5}$
c) $\frac{3}{4} \times \left( {\frac{5}{2} - \frac{3}{2}} \right) = \frac{3}{4} \times 1 = \frac{3}{4}$
\(\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}=\dfrac{1}{5}\)
\(0,75-\left(2\dfrac{1}{3}+0,75\right)+3^2\times\left(-\dfrac{1}{9}\right)\)
\(=0,75-2\dfrac{1}{3}-0,75+9\times\left(-\dfrac{1}{9}\right)\)
\(=\left(0,75-0,75\right)-\left(2+\dfrac{1}{3}\right)+\left(-\dfrac{9}{9}\right)\)
\(=0-2-\dfrac{1}{3}-1\)
\(=-3-\dfrac{1}{3}\)
\(=-\dfrac{10}{3}\)