\(\frac{1}{3}\times\frac{1}{6}\times\frac{1}{9}=\frac{1\times1\times1}{3\times6\times9}=\frac{1}{162}\)

\(\frac{1}{3}\times\frac{1}{6}:\frac{1}{9}=\frac{1}{3}\times\frac{1}{6}\times\frac{9}{1}=\frac{1\times1\times9}{3\times6\times1}=\frac{9}{18}=\frac{1}{2}\)

Tự tính câu cuối đi

a) 2/9 : 2/3 x 1/2

= 2/9 x 3/2 x 1/2

= 6/18 x 1/2

= 6/36

b) 2 + 1/4 x 4/3

= 2 + 4/7

= 2/1₍₇₎ + 4/7

= 14/7 + 4/7

= 18/7

c) 3 x 1/2 x 1/4

= 3/1 x 1/2 x 1/4

= 3/2 x 1/4

= 3/8

#Hok_tốt

Thông báo,Te2qAMq.png (1280×800)

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\(\frac{1}{1x3}+\frac{1}{3x5}+...+\frac{1}{95x97}+\frac{1}{97x99}\)

\(=\frac{1}{2}x\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+,,,+\frac{1}{97}-\frac{1}{99}\right)\)

\(=\frac{1}{2}x\frac{98}{99}\)

\(=\frac{49}{99}\)

Đặt A=1/3x1 + 1/3x5 + ......+ 1/95x97 + 1/97x99

\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\)

\(2A=1-\frac{1}{99}\)

\(A=\frac{98}{99}:2\)

\(A=\frac{49}{99}\)

sao lại làm như vậy

1/3x1/9-1/10=---17/270

hình như bạn ra sai đề đó tính ra bằng âm

\(\frac{1}{3}\times\frac{1}{9}-\frac{1}{10}=\frac{1}{27}-\frac{1}{10}=-\frac{17}{270}\)

1/3x1/9=1/27-1/10=17/270

\(\frac{1}{2}\)- \(\frac{1}{3}\)x \(\frac{1}{4}\)

= \(\frac{1}{2}\)- \(\frac{1}{12}\)

= \(\frac{6}{12}\)- \(\frac{1}{12}\)

= \(\frac{5}{12}\)

tk nhé

5

12

 kb và k nha

1/2-1/3x1/4=1/24

k cho mình nhé

\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{9999}=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)

\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{101}\right)=\frac{1}{2}.\frac{100}{101}=\frac{50}{101}\)

\(\frac{1}{2}-\frac{1}{3}\cdot\frac{1}{4}\)

=\(\frac{1}{2}-\frac{1}{3.4}\)

=\(\frac{1}{2}-\frac{1}{12}\)

=\(\frac{6}{12}-\frac{1}{12}\)

=\(\frac{6-1}{12}\)

=\(\frac{5}{12}\)

Đặt \(A=\frac{1}{99.97}-\frac{1}{97.95}-\frac{1}{95.93}-....-\frac{1}{5.3}-\frac{1}{3.1}\)

\(\Rightarrow A=\frac{1}{99.97}-\left(\frac{1}{1.3}+\frac{1}{3.5}+....+\frac{1}{93.95}+\frac{1}{95.97}\right)\)

\(\Rightarrow A=\frac{1}{99.97}-\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{95}-\frac{1}{97}\right)\)

\(\Rightarrow A=\frac{1}{99}-\frac{1}{97}-\frac{1}{2}\left(1-\frac{1}{97}\right)=\frac{1}{99}-\frac{1}{97}-\frac{1}{2}-\frac{1}{194}\)