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\(3M=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\)
\(3M=\frac{4-1}{1.4}+\frac{7-4}{4.7}+...+\frac{100-97}{97.100}\)
\(3M=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\)
\(3M=1-\frac{1}{100}\)
\(3M=\frac{99}{100}\)
\(M=\frac{33}{100}\)
\(A=\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+\frac{1}{130}+...+\frac{1}{9700}\)
\(A=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{97.100}\)
\(A=\frac{3}{3}\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{97.100}\right)\)
\(A=\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{97.100}\right)\)
\(A=\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(A=\frac{1}{3}\left(1-\frac{1}{100}\right)\)
\(A=\frac{1}{3}.\frac{99}{100}=\frac{33}{100}\)
\(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+\frac{1}{130}+...+\frac{1}{9700}\)
\(=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}\)
\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\)
\(=\frac{1}{1}-\frac{1}{100}\)
\(=\frac{100}{100}-\frac{1}{100}\)
\(=\frac{99}{100}\)
\(\frac{3}{1.4}+\frac{3}{4.7}+..+\frac{3}{97.100}=\frac{0,33x}{2009}\)
\(1-\frac{1}{4}+\frac{1}{4}-...-\frac{1}{100}=\frac{0,33x}{2009}\)
\(1-\frac{1}{100}=\frac{0,33x}{2009}\)
\(\frac{99}{100}=\frac{0,33x}{20009}\Rightarrow2009.99=100.0,33x\)
x=6027
$\dfrac14+\dfrac1{28}+\dfrac1{70}+\cdots+\dfrac1{9700}$
$=\dfrac1{2\cdot2}+\dfrac1{4\cdot7}+\dfrac1{7\cdot10}+\cdots+\dfrac1{97\cdot100}$
$=\dfrac13\left(\dfrac12-\dfrac15\right)+\dfrac13\left(\dfrac14-\dfrac17\right)+\dfrac13\left(\dfrac17-\dfrac1{10}\right)+\cdots+\dfrac13\left(\dfrac1{97}-\dfrac1{100}\right)$
$=\dfrac13\left(\dfrac12-\dfrac15+\dfrac14-\dfrac17+\dfrac17-\dfrac1{10}+\cdots+\dfrac1{97}-\dfrac1{100}\right)$
$=\dfrac13\left(\dfrac12+\dfrac14-\dfrac15-\dfrac1{100}\right)$
$=\dfrac13\cdot\dfrac{54}{100}$
$=\dfrac9{50}$.
Theo đề bài:
$\dfrac9{50}=\dfrac{0,33x}{2009}$
$\Leftrightarrow\dfrac9{50}=\dfrac{33x}{100\cdot2009}$
$\Leftrightarrow 9\cdot100\cdot2009=50\cdot33x$
$\Leftrightarrow x=\dfrac{9\cdot100\cdot2009}{50\cdot33}$
$\Leftrightarrow x=\dfrac{6\cdot2009}{11}$
$\Leftrightarrowx=\dfrac{12054}{11}$.
\(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+...+\frac{1}{9700}=\frac{0,33x}{2009}\)
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}=\frac{0,33x}{2009}\)
\(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}=\frac{0,33x}{2009}\)
\(\frac{1}{1}-\frac{1}{100}=\frac{0,33x}{2009}\)
\(\frac{100}{100}-\frac{1}{100}=\frac{0,33x}{2009}\)
\(\frac{99}{100}=\frac{0,33x}{2009}\)
\(\Rightarrow2009.99=100.0,33x\)
\(\Rightarrow2009.99=33x\)
\(\Rightarrow2009.99:33=x\)
\(\Rightarrow2009.3=x\)
\(\Rightarrow6027=x\)
Vậy \(x=6027\)(MK KO CHẮC NÓ ĐÚNG NHÉ )
A = 1/4 + 1/28 + 1/70 +...+ 1/9700
A = 1/1.4 + 1/4.7 + 1/7.10 +...+ 1/97.100
3A = 3/1.4 + 3/4.7 + 3/7.10 +...+ 3/97.100
3A = 1 - 1/100
3A = 99/100
A=99/100:3=33/100
\(=\frac{1}{1.4}+\frac{1}{4.7}+..+\frac{1}{97.100}\)
\(=\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{97.100}\right)\)
\(=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{100}\right)\)
\(=\frac{1}{3}.\frac{99}{100}=\frac{33}{100}\)
ta có:
\(M=\frac14+\frac{1}{28}+\frac{1}{70}+\cdots+\frac{1}{9700}\)
\(M=\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+\cdots+\frac{1}{97\cdot100}\)
\(M=1-\frac14+\frac14-\frac17+\frac17-\frac{1}{10}+\cdots+\frac{1}{97}-\frac{1}{100}\)
\(M=1-\frac{1}{100}\)
\(M=\frac{99}{100}\)
nhanh plzz em đang vội ai xong trc em tích
\(M=\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+\ldots+\frac{1}{9700}\)
\(M=\frac{1}{1 \cdot4}+\frac{1}{4 \cdot7}+\frac{1}{7 \cdot10}+\ldots+\frac{1}{97 \cdot100}\)
\(M=\frac{1}{3}\left(\right.\frac{3}{1 \cdot4}+\frac{3}{4 \cdot7}+\frac{3}{7 \cdot10}+\ldots+\frac{3}{97 \cdot100}\left.\right)\)
\(M=\frac{1}{3}\left[\right.\left(\right.1-\frac{1}{4}\left.\right)+\left(\right.\frac{1}{4}-\frac{1}{7}\left.\right)+\left(\right.\frac{1}{7}-\frac{1}{10}\left.\right)+\ldots+\left(\right.\frac{1}{97}-\frac{1}{100}\left.\right)\left]\right.\)
\(M=\frac{1}{3}\left(\right.1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\ldots+\frac{1}{97}-\frac{1}{100}\left.\right)\)
\(M = \frac{1}{3} \left(\right. 1 - \frac{1}{100} \left.\right)\)
\(M=\frac{1}{3}\times\frac{99}{100}\)
\(M=\frac{33}{100}\)
`x/(a*b) = 1/a - 1/b <=> b-a = x`
`M = 1/4 + 1/28 + 1/70 + 1/130 + ... + 1/9700`
`=> M = 1/(1*4) + 1/(4*7) + ... + 1/(97*100)`
`=> 3M = 3/(1*4) + 3/(4*7) + ...+ 3/(97*100)`
`=> 3M = 1 - 1/4 + 1/4 - 1/7 +...+ 1/97 - 1/100`
`=> 3M = 1 - 1/100`
`=> 3M = 99/100`
`=> M = 99/100 : 3 = 33/100`
Vậy `M=33/100`
`x/(a*b) = 1/a - 1/b <=> b-a=x`
Mà `4-1 =3 ` nên `1/(1*4)≠1-1/4`