1/5x6+1/6x7+1/7x8+...+1/2019x2020
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\(\frac{1}{5×6}+\frac{1}{6×7}+\frac{1}{7×8}+...+\frac{1}{24×25}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}\)
\(=\frac{1}{5}-\frac{1}{25}\)
\(=\frac{5}{25}-\frac{1}{25}\)
\(=\frac{4}{25}\)
\(\frac{1}{5\times6}+\frac{1}{6\times7}+\frac{1}{7\times8}+...+\frac{1}{24\times25}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}\)
\(=\frac{1}{5}-\frac{1}{25}\)
\(=\frac{4}{25}\)\(\left(=0,16\right)\)
1/5×6 + 1/6×7 + 1/7×8 + ... + 1/24×25
= 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 + ... + 1/24 - 1/25
= 1/5 - 1/25
= 5/25 - 1/25
= 4/25
\(A=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{7}-\dfrac{1}{8}=\dfrac{1}{2}-\dfrac{1}{8}=\dfrac{3}{8}\)
Đặt \(A=\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}+\frac{1}{6x7}+\frac{1}{7x8}+\frac{1}{8x9}\)
\(A=\frac{3-2}{2x3}+\frac{4-3}{3x4}+\frac{5-4}{4x5}+\frac{6-5}{5x6}+\frac{7-6}{6x7}+\frac{8-7}{7x8}+\frac{9-8}{8x9}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)
\(A=\frac{1}{2}-\frac{1}{9}=\frac{9}{18}-\frac{2}{18}=\frac{7}{18}\)
= 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9 + 1/9 - 1/10 + 1/10 - 1/11 + 1/11 - 1/12
= 1/3 - 1/12 = 1/4
Tra loi:
1/2x3+1/3x4+1/4x5+1/5x6+1/6x7+1/7x8+1/8x9
=1/2 - 1/3 + 1/3 - 1/4 +1/4 - 1/5 + 1/5 -1/6 +1/6 - 1/7 +1/7 -1/8+1/8 - 1/9
=1/2-(1/3-1/3)-(1/4-1/4)-(1/5-1/5)-(1/6-1/6)-(1/7-1/7)-(1/8-1/8)-1/9
=1/2-0-0-0-0-0-0-1/9
=1/2-1/9
=7/18
k cho minh nha!
thông điệp nhỏ
hai tích nếu như ko muốn k
ai tích mình tích lại
1/5x6+1/6x7+1/7x8+...+1/2019x2020
= 6/5+7/6+8/7+...+2020/2019
Rút gọn cho nhau ta còn 2020/5=404
1/5 x 6 + 1/6 x7 + 1/7 x8 + ... + 1/2019 - 1/ 2020
=1/5 -1/6 +1/6 -1/7 + 1/7 - 1/8 + ... + 1/2019 - 1/2020
Sau khi giản ước, ta còn:
1/5 - 1/2020 = 403/2020.
Đáp số: 403/2020