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Từ bé đến lớn : 13/14;14/15;15/16;16/17;17/18;18/19;19/20
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\(\dfrac{1}{3\times7}+\dfrac{1}{7\times11}+\dfrac{1}{11\times15}+...+\dfrac{1}{a\times\left(a+4\right)}=\dfrac{50}{609}\)
\(\dfrac{1}{4}\times\left(\dfrac{4}{3\times7}+\dfrac{4}{7\times11}+...+\dfrac{4}{a\times\left(a+4\right)}\right)=\dfrac{50}{609}\)
\(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{a}-\dfrac{1}{a\times4}=\dfrac{50}{609}\div\dfrac{1}{4}\)
\(\dfrac{1}{3}-\dfrac{1}{a\times4}=\dfrac{200}{609}\)
\(\dfrac{1}{a\times4}=\dfrac{1}{3}-\dfrac{200}{609}\)
\(\dfrac{1}{a\times4}=\dfrac{1}{203}\)
\(a\times4=203\)
\(a=\dfrac{203}{4}\)
\(\dfrac{1}{3\times7}\)+\(\dfrac{1}{7\times11}\)+\(\dfrac{1}{11\times15}\)+...+\(\dfrac{1}{a\times\left(a+4\right)}\) = \(\dfrac{50}{609}\)
4\(\times\)( \(\dfrac{1}{3\times7}\) +\(\dfrac{1}{7\times11}\)+\(\dfrac{1}{11\times15}\)+...+\(\dfrac{1}{a\times\left(a+4\right)}\)) = \(\dfrac{50}{609}\) \(\times\)4
\(\dfrac{4}{3\times7}\)+ \(\dfrac{4}{7\times11}\)+\(\dfrac{1}{11\times15}\)+...+\(\dfrac{4}{a\times\left(a+4\right)}\) = \(\dfrac{50}{609}\) \(\times\) 4
\(\dfrac{1}{3}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{11}\) + \(\dfrac{1}{11}\)-\(\dfrac{1}{15}\)+...+\(\dfrac{1}{a}\)-\(\dfrac{1}{a+4}\) = \(\dfrac{200}{609}\)
\(\dfrac{1}{3}\) - \(\dfrac{1}{a+4}\) = \(\dfrac{200}{609}\)
\(\dfrac{1}{a+4}\) = \(\dfrac{1}{3}\) - \(\dfrac{200}{609}\)
\(\dfrac{1}{a+4}\) = \(\dfrac{1}{203}\)
a + 4 = 203
\(a\) = 203 - 4
\(a\) = 199
Đáp số: \(a\) = 199
Cách 1 :
Ta có : \(\frac{11}{12}=\frac{198}{216}\)
\(\frac{17}{18}=\frac{201}{216}\)
Vì : 198 < 201 => \(\frac{198}{216}=\frac{11}{12}< \frac{201}{216}=\frac{17}{18}\)
Cách 2 :
Có : \(\frac{11}{12}=1-\frac{1}{12}\)
\(\frac{17}{18}=1-\frac{1}{18}\)
Vì : \(\frac{1}{12}>\frac{1}{18}\)=>\(\frac{11}{12}< \frac{17}{18}\)
Ủng hộ nhá !!!
<tfh
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\(A=\frac{18}{11}+\frac{11}{13}+\frac{15}{17}+\frac{17}{15}\)
\(=1+\frac{7}{11}+1-\frac{2}{13}+1-\frac{2}{17}+1+\frac{2}{15}\)
\(=4+\frac{7}{11}+\frac{2}{15}-\frac{2}{13}-\frac{2}{17}\)
\(\frac{7}{11}>\frac{2}{11}>\frac{2}{13};\frac{2}{15}>\frac{2}{17}\Rightarrow\frac{7}{11}+\frac{2}{15}>\frac{2}{11}+\frac{2}{17}\)
\(\Rightarrow\frac{7}{11}+\frac{2}{15}-\frac{2}{13}-\frac{2}{17}>0\)
\(\Rightarrow A>4+0=4\)
Vậy A>4