1/20+1/30+1/42+...1/9900
/ là phân số nha!
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\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)
\(=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{99\times100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
1/2 + 1/6 + 1/12 + ... + 1/9900
1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +... + 1/99 - 1/100
1/1 - 1/100
= 99/100
Ta có \(\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{9900}\right):x=\frac{1}{5}\)
\(\left(\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+\frac{1}{6\times7}+...+\frac{1}{99\times100}\right):x=\frac{1}{5}\)
\(\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\right):x=\frac{1}{5}\)
\(\left(\frac{1}{3}-\frac{1}{100}\right):x=\frac{1}{5}\)
\(\frac{97}{300}:x=\frac{1}{5}\)
\(x=\frac{97}{300}:\frac{1}{5}\)
\(x=\frac{97}{60}\)
\(A=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+...+\frac{1}{9900}\)
\(A=\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+...+\frac{1}{99\cdot100}\)
\(A=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{4}-\frac{1}{100}\)
\(A=\frac{6}{25}\)
\(A=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+...+\frac{1}{9900}\)
\(=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+.....+\frac{1}{99.100}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+....+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{4}-\frac{1}{100}\\ =\frac{24}{100}=\frac{6}{25}\)
\(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+...+\dfrac{1}{50}+\dfrac{1}{990}\)
\(=\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{50}+\dfrac{1}{990}???\)
Quy luật của vế sau "..." sai, bạn xem lại đề bài!
Nếu đúng đề thì sẽ như sau:
\(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+...+\dfrac{1}{9900}\)
Đề bài đúng là như vậy.
Giải:
\(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+...+\dfrac{1}{9900}\)
\(=\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\dfrac{1}{4}-\dfrac{1}{100}\)
\(=\dfrac{25-1}{100}\)
\(=\dfrac{24}{100}\)
\(=\dfrac{6}{25}\)
\(A=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{9900}\)
\(A=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+....+\frac{1}{99.100}\)
\(A=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+....+\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{4}-\frac{1}{100}=\frac{6}{25}\)
Vậy A=6/25
\(A=\frac{1}{4\times5}+\frac{1}{5\times6}+\frac{1}{6\times7}+.....+\frac{1}{99\times100}\)
\(A=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+.....+\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{4}-\frac{1}{100}\)
\(A=\frac{24}{100}=\frac{6}{25}\)
\(A=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{9900}\)
\(A=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}-\frac{1}{6}+\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{4}-\frac{1}{100}\)
\(A=\frac{6}{25}\)
\(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{9900}\)
\(=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+....+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{4}-\frac{1}{100}\)
\(=\frac{25-1}{100}=\frac{24}{100}=\frac{6}{25}\)
\(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{9900}\)
\(=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{4}-\frac{1}{100}\)
\(=\frac{6}{25}\)
k mình nha !