A=\(\dfrac{2}{15}\)+\(\dfrac{2}{35}\)+\(\dfrac{2}{63}\)+...+\(\dfrac{2}{399}\)
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Những câu hỏi liên quan
KM
2
15 tháng 5 2021
Giải:
\(\dfrac{4}{15}+\dfrac{4}{35}+\dfrac{4}{63}+...+\dfrac{4}{399}=\dfrac{x}{49}\)
\(\dfrac{4}{3.5}+\dfrac{4}{5.7}+\dfrac{4}{7.9}+...+\dfrac{4}{19.21}=\dfrac{x}{49}\)
\(2.\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{19.21}\right)=\dfrac{x}{49}\)
\(2.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{19}-\dfrac{1}{21}\right)=\dfrac{x}{49}\)
\(2.\left(\dfrac{1}{3}-\dfrac{1}{21}\right)=\dfrac{x}{49}\)
\(2.\dfrac{2}{7}=\dfrac{x}{49}\)
\(\dfrac{4}{7}=\dfrac{x}{49}\)
\(\Rightarrow x=\dfrac{4.49}{7}=28\)
Chúc bạn học tốt!
15 tháng 5 2021
\(\dfrac{4}{15}+\dfrac{4}{35}+...+\dfrac{4}{399}=\dfrac{x}{49}\)
2 . \(\dfrac{2}{15}+\dfrac{2}{35}+...+\dfrac{2}{399}=\dfrac{x}{49}\)
2 . \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{19.21}=\dfrac{x}{49}\)
2 . ( \(\dfrac{1}{3}-\dfrac{1}{21}\) ) = \(\dfrac{x}{49}\)
2 . \(\dfrac{2}{7}\) = \(\dfrac{x}{49}\)
=> \(\dfrac{4}{7}=\dfrac{x}{49}\)
=> \(\dfrac{21}{49}=\dfrac{x}{49}\)
=> \(x=21\)
Vậy \(x=21\)
15 tháng 5 2021
Ta có : \(\dfrac{4}{15}+\dfrac{4}{35}+\dfrac{4}{63}+...+\dfrac{4}{399}=\dfrac{x}{49}\)
\(\Leftrightarrow2\cdot\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{19.21}\right)=\dfrac{x}{49}\)
\(\Leftrightarrow\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{19}-\dfrac{1}{21}=\dfrac{x}{98}\)
\(\Leftrightarrow\dfrac{1}{3}-\dfrac{1}{21}=\dfrac{x}{98}\)
\(\Leftrightarrow\dfrac{2}{7}=\dfrac{x}{98}\Rightarrow x=28\)
Vậy $x=28$
17 tháng 7 2017
\(C=\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+...+\dfrac{2}{399}\)
\(C=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{19.21}\)
\(C=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{19}-\dfrac{1}{21}\)
\(C=\dfrac{1}{3}-\dfrac{1}{21}\)
\(C=\dfrac{2}{7}\)
14 tháng 6 2021
Giải:
\(\dfrac{2}{3}+\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+\dfrac{2}{99}+\dfrac{2}{143}\)
\(=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\)
\(=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\)
\(=\dfrac{1}{1}-\dfrac{1}{13}\)
\(=\dfrac{12}{13}\)
Chúc em học tốt!
14 tháng 6 2021
2/3+2/15+2/35+2/63+2/99+2/143
=2(1/1x3+1/3x5+1/5x7+1/7x9+1/9x11+1/11x13)
=2(1-1/3+1/3-1/5+1/5-....+1/13)
=2(1-1/13)
=2.12/13=24/13
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
23 tháng 3 2023
( \(\dfrac{2}{15}\) + \(\dfrac{2}{35}\) + \(\dfrac{2}{63}\)) : \(x\) = \(\dfrac{1}{18}\)
( \(\dfrac{2\times7}{15\times7}\) + \(\dfrac{2\times3}{35\times3}\) + \(\dfrac{2}{63}\)) : \(x\) = \(\dfrac{1}{18}\)
(\(\dfrac{14}{105}\) + \(\dfrac{6}{105}\) + \(\dfrac{2}{63}\)) : \(x\) = \(\dfrac{1}{18}\)
(\(\dfrac{20}{105}\) + \(\dfrac{2}{63}\)) : \(x\) = \(\dfrac{1}{18}\)
( \(\dfrac{4}{21}\) + \(\dfrac{2}{63}\)) : \(x\) = \(\dfrac{1}{18}\)
(\(\dfrac{12}{63}\) + \(\dfrac{2}{63}\)) : \(x\) = \(\dfrac{1}{18}\)
\(\dfrac{2}{9}\) : \(x\) = \(\dfrac{1}{18}\)
\(x\) = \(\dfrac{2}{9}\) : \(\dfrac{1}{18}\)
\(x\) = 4
DT
1
8 tháng 7 2017
\(A=\dfrac{14}{8.11}+\dfrac{14}{11.14}+\dfrac{14}{14.17}+.....+\dfrac{14}{197.200}\)
\(A=\dfrac{14}{3}\left(\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}+...+\dfrac{1}{197}-\dfrac{1}{200}\right)\)
\(A=\dfrac{14}{3}.\left(\dfrac{1}{8}-\dfrac{1}{200}\right)\)
\(A=\dfrac{14}{3}.\dfrac{24}{200}=\dfrac{28}{25}\)
\(B=\dfrac{7}{15}+\dfrac{7}{35}+\dfrac{7}{63}+...+\dfrac{7}{399}\)
\(B=\dfrac{7}{3.5}+\dfrac{7}{5.7}+\dfrac{7}{7.9}+.....\dfrac{7}{19.21}\)
\(B=\dfrac{7}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+....+\dfrac{1}{19}-\dfrac{1}{21}\right)\)
\(B=\dfrac{7}{2}.\left(\dfrac{1}{3}-\dfrac{1}{21}\right)\)
\(B=\dfrac{7}{2}.\dfrac{6}{21}=1\)
AH
Akai Haruma
Giáo viên
30 tháng 4 2022
Lời giải:
Vế trái luôn không âm (tính chất trị tuyệt đối)
$\Rightarrow -11x\geq 0$
$\Rightarrow x\leq 0$
Do đó: $x-\frac{1}{3}, x-\frac{1}{15},..., x-\frac{1}{399}<0$
PT trở thành:
$\frac{1}{3}-x+\frac{1}{15}-x+...+\frac{1}{399}-x=-11x$
$(\frac{1}{3}+\frac{1}{15}+...+\frac{1}{399})-10x=-11x$
$\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{19.21}=-x$
$\frac{1}{2}(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{19}-\frac{1}{21})=-x$
$\frac{1}{2}(1-\frac{1}{21})=-x$
$\frac{10}{21}=-x$
$\Rightarrow x=\frac{-10}{21}$
AH
Akai Haruma
Giáo viên
30 tháng 4 2022
Lời giải:
Vế trái luôn không âm (tính chất trị tuyệt đối)
$\Rightarrow -11x\geq 0$
$\Rightarrow x\leq 0$
Do đó: $x-\frac{1}{3}, x-\frac{1}{15},..., x-\frac{1}{399}<0$
PT trở thành:
$\frac{1}{3}-x+\frac{1}{15}-x+...+\frac{1}{399}-x=-11x$
$(\frac{1}{3}+\frac{1}{15}+...+\frac{1}{399})-10x=-11x$
$\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{19.21}=-x$
$\frac{1}{2}(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{19}-\frac{1}{21})=-x$
$\frac{1}{2}(1-\frac{1}{21})=-x$
$\frac{10}{21}=-x$
$\Rightarrow x=\frac{-10}{21}$
`A =2/15 +2/35 +2/63 +... +2/339`
`= 2/(3.5) +2/(5.7) + 2/(7.9) + ...+2/(19.21)`
`= 1/3 -1/5 +1/5 -1/7 +1/7 -1/9 +... 1/19 -1/21`
`= 1/3 -1/21 = 7/21 -1/21`
`=6/21 = 2/7`
A=2/15+2/35+2/63+...+2/339
=2/(3.5)+2/(5.7)+2/(7.9)+...+2/(19.21)=2/(3.5)+2/(5.7)+2/(7.9)+...+2/(19.21)
=1/3−1/5+1/5−1/7+1/7−1/9+...1/19−1/21=1/3−1/5+1/5−1/7+1/7−1/9+...1/19−1/21
=1/3−1/21=7/21−1/21=1/3−1/21=7/21−1/21
=6/21=2/7=6/21=2/7