Tính :
\(\dfrac{4}{15}\) + \(\dfrac{4}{35}\) + \(\dfrac{4}{63}\) + ... + \(\dfrac{4}{399}\)
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15 tháng 5 2021 lúc 15:28
Tìm x :
\(\dfrac{4}{15}+\dfrac{4}{35}+\dfrac{4}{63}+...+\dfrac{4}{399}=\dfrac{x}{49}\)
Giải:
\(\dfrac{4}{15}+\dfrac{4}{35}+\dfrac{4}{63}+...+\dfrac{4}{399}=\dfrac{x}{49}\)
\(\dfrac{4}{3.5}+\dfrac{4}{5.7}+\dfrac{4}{7.9}+...+\dfrac{4}{19.21}=\dfrac{x}{49}\)
\(2.\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{19.21}\right)=\dfrac{x}{49}\)
\(2.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{19}-\dfrac{1}{21}\right)=\dfrac{x}{49}\)
\(2.\left(\dfrac{1}{3}-\dfrac{1}{21}\right)=\dfrac{x}{49}\)
\(2.\dfrac{2}{7}=\dfrac{x}{49}\)
\(\dfrac{4}{7}=\dfrac{x}{49}\)
\(\Rightarrow x=\dfrac{4.49}{7}=28\)
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\(\dfrac{4}{15}+\dfrac{4}{35}+...+\dfrac{4}{399}=\dfrac{x}{49}\)
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 \(x=21\)
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15 tháng 5 2021 lúc 15:24
Tìm x :
\(\dfrac{4}{15}\) + \(\dfrac{4}{35}\) + \(\dfrac{4}{63}\) + ... + \(\dfrac{4}{399}\) = \(\dfrac{x}{49}\)
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 $x=28$
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Thực hiện phép tính
a) Aleft(dfrac{dfrac{4}{15}+dfrac{4}{35}+dfrac{4}{63}+.......+dfrac{4}{399}}{dfrac{3^2}{8.11}+dfrac{3^2}{11.14}+dfrac{3^2}{14.17}+.....+dfrac{3^2}{197.200}}right).dfrac{200720072007}{200820082008}
b) B1.sqrt{2}+2.sqrt{3}+3.sqrt{4}+....+9sqrt{10}
c) D dfrac{2006}{0,20072008...}+dfrac{2007}{0,020072008...}+dfrac{2008}{0,0020072008}
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Thực hiện phép tính
a) A=\(\left(\dfrac{\dfrac{4}{15}+\dfrac{4}{35}+\dfrac{4}{63}+.......+\dfrac{4}{399}}{\dfrac{3^2}{8.11}+\dfrac{3^2}{11.14}+\dfrac{3^2}{14.17}+.....+\dfrac{3^2}{197.200}}\right).\dfrac{200720072007}{200820082008}\)
b) B=\(1.\sqrt{2}+2.\sqrt{3}+3.\sqrt{4}+....+9\sqrt{10}\)
c) D = \(\dfrac{2006}{0,20072008...}+\dfrac{2007}{0,020072008...}+\dfrac{2008}{0,0020072008}\)
1 tháng 3 2022 lúc 16:27
Tính: \(A=\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}.\dfrac{24}{25}.\dfrac{35}{36}.\dfrac{48}{49}.\dfrac{63}{64}\)
A=\(\dfrac{2}{15}\)+\(\dfrac{2}{35}\)+\(\dfrac{2}{63}\)+...+\(\dfrac{2}{399}\)
`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`
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sai rồi,2/7 mới đúng,bài này không cần nhân 2
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A=\(\dfrac{2}{3}\)+\(\dfrac{14}{15}\)+\(\dfrac{34}{35}\)+\(\dfrac{62}{63}\)+\(\dfrac{98}{99}\)+\(\dfrac{142}{143}\)+\(\dfrac{194}{195}\)
Và B=5+\(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^3}\)+\(^{\dfrac{1}{4^4}}\)+\(\dfrac{1}{5^5}\)+\(\dfrac{1}{6^6}\)+\(\dfrac{1}{7^7}\).So sánh A và B
tìm x biết \(\left|x-\dfrac{1}{3}\right|+\left|x-\dfrac{1}{15}\right|+\left|x-\dfrac{1}{35}\right|+\left|x-\dfrac{1}{63}\right|+...+\left|x-\dfrac{1}{399}\right|=-11x\)
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:
$\frac{1}{3}-x+\frac{1}{15}-x+...+\frac{1}{399}-x=-11x$
$(\frac{1}{3}+\frac{1}{15}+...+\frac{1}{399})-10x=-11x$
$\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{19.21}=-x$
$\frac{1}{2}(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{19}-\frac{1}{21})=-x$
$\frac{1}{2}(1-\frac{1}{21})=-x$
$\frac{10}{21}=-x$
$\Rightarrow x=\frac{-10}{21}$
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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:
$\frac{1}{3}-x+\frac{1}{15}-x+...+\frac{1}{399}-x=-11x$
$(\frac{1}{3}+\frac{1}{15}+...+\frac{1}{399})-10x=-11x$
$\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{19.21}=-x$
$\frac{1}{2}(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{19}-\frac{1}{21})=-x$
$\frac{1}{2}(1-\frac{1}{21})=-x$
$\frac{10}{21}=-x$
$\Rightarrow x=\frac{-10}{21}$
\(\dfrac{4}{15}\)+\(\dfrac{4}{35}\)+.....+\(\dfrac{4}{195}\)+\(\dfrac{4}{323}\)
\(=2\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{17.19}\right)\)
\(=2\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{17}-\dfrac{1}{19}\right)\)
\(=2\left(\dfrac{1}{3}-\dfrac{1}{19}\right)=2\text{ }\times\dfrac{16}{57}=\dfrac{32}{57}\)
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1. Tính
a. \(A=\dfrac{14}{8.11}+\dfrac{14}{11.14}+\dfrac{14}{14.17}+...+\dfrac{14}{197.200}\)
b. \(B=\dfrac{7}{15}+\dfrac{7}{35}+\dfrac{7}{63}+...+\dfrac{7}{399}\)
\(A=\dfrac{14}{8.11}+\dfrac{14}{11.14}+\dfrac{14}{14.17}+.....+\dfrac{14}{197.200}\)
\(A=\dfrac{14}{3}\left(\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}+...+\dfrac{1}{197}-\dfrac{1}{200}\right)\)
\(A=\dfrac{14}{3}.\left(\dfrac{1}{8}-\dfrac{1}{200}\right)\)
\(A=\dfrac{14}{3}.\dfrac{24}{200}=\dfrac{28}{25}\)
\(B=\dfrac{7}{15}+\dfrac{7}{35}+\dfrac{7}{63}+...+\dfrac{7}{399}\)
\(B=\dfrac{7}{3.5}+\dfrac{7}{5.7}+\dfrac{7}{7.9}+.....\dfrac{7}{19.21}\)
\(B=\dfrac{7}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+....+\dfrac{1}{19}-\dfrac{1}{21}\right)\)
\(B=\dfrac{7}{2}.\left(\dfrac{1}{3}-\dfrac{1}{21}\right)\)
\(B=\dfrac{7}{2}.\dfrac{6}{21}=1\)
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