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Ta có : \(\frac{1}{1\times2}+\frac{1}{2\times3}+....+\frac{1}{99\times100}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
= \(1-\frac{1}{100}=\frac{99}{100}\)
Bài này lớp 6 phải không bạn
A=1/1-1/2+1/2-1/3+1/3-1/4+1/5-1/6+......................+1/99-1/100
A=1/1-1/100
A=99/100
Nếu bạn cảm thấy bài mình đúng thì cho mình một "lai"
A = 5(1/1.2 + 1/2.3 +......+ 1/99.100)
A = 5( 1 - 1/2 + 1/2 - 1/3 +........+ 1/99 - 1/100)
A = 5( 1 - 1/100)
A = 5 . 99/100
A = 99/20
** k mk nha!
\(\frac{5}{1\times2}+\frac{5}{2\times3}+...+\frac{5}{99\times100}=5\left(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{99\times100}\right)=5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)=5\left(1-\frac{1}{100}\right)=5\times\frac{99}{100}=\frac{99}{20}=4\frac{19}{20}\)
\(\Rightarrow A=5\left(\frac{1}{1x2}+\frac{1}{2x3}+...+\frac{1}{99x100}\right)\)
\(\Rightarrow A=5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(\Rightarrow A=5\left(1-\frac{1}{100}\right)\)
\(\Rightarrow A=\frac{5x99}{100}=\frac{99}{20}\)
\(A=\frac{5}{1}-\frac{5}{2}+\frac{5}{2}-\frac{5}{3}+\frac{5}{3}-\frac{5}{4}+....+\frac{5}{99}-\frac{5}{100}\)
\(A=\frac{5}{1}+\left(-\frac{5}{2}+\frac{5}{2}\right)+\left(-\frac{5}{3}+\frac{5}{3}\right)+\left(-\frac{5}{4}+\frac{5}{4}\right)+...\left(-\frac{5}{99}+\frac{5}{99}\right)+\frac{5}{100}\)
\(A=\frac{5}{1}+0+0+....+0+\frac{5}{100}\)
\(A=\frac{500}{100}+\frac{5}{100}=\frac{205}{100}=\frac{101}{20}\)
Đúng 100%
Đúng 100%
Đúng 100%
\(=\frac{1.2}{99.100}\)
\(=\frac{2}{9900}=\frac{1}{4950}\)
Ta có:
\(A=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}...\frac{99^2}{99.100}.\frac{100^2}{100.101}\)
\(=\frac{1}{2}.\frac{4}{6}.\frac{9}{12}....\frac{9801}{9900}.\frac{10000}{10100}\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{99}{100}.\frac{100}{101}=\frac{1.2.3...99.100}{2.3.4...100.101}=\frac{1}{101}\)(Tối giản)
\(a=\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{99\times100}\)
\(a=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(a=1-\frac{1}{100}=\frac{100-1}{100}=\frac{99}{100}\)
Vậy \(a=\frac{99}{100}\)
Ta có : \(\frac{\frac{3}{5}+\frac{3}{7}-\frac{1}{3}+\frac{3}{11}}{\frac{6}{5}+\frac{6}{7}-\frac{2}{3}+\frac{6}{11}}=\frac{\frac{3}{5}+\frac{3}{7}-\frac{1}{3}+\frac{3}{11}}{2\left(\frac{3}{5}+\frac{3}{7}-\frac{1}{3}+\frac{3}{11}\right)}=\frac{1}{2}\)
Lại có : \(\frac{\left(\frac{1}{4}-\frac{1}{5}-\frac{1}{20}\right).2021}{\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}}=\frac{0.2021}{\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}}=0\)
Khi đó \(B=\frac{1}{2}+0=\frac{1}{2}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
\(A=\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{99\times100}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)