Tìm phân số \(\frac{a}{b}\) biết \(\frac{11}{17}\)< \(\frac{a}{b}\)< \(\frac{23}{29}\)
và 8a - 9b = 31
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\dfrac{1}{3\times7}+\dfrac{1}{7\times11}+\dfrac{1}{11\times15}+...+\dfrac{1}{a\times\left(a+4\right)}=\dfrac{50}{609}\)
\(\dfrac{1}{4}\times\left(\dfrac{4}{3\times7}+\dfrac{4}{7\times11}+...+\dfrac{4}{a\times\left(a+4\right)}\right)=\dfrac{50}{609}\)
\(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{a}-\dfrac{1}{a\times4}=\dfrac{50}{609}\div\dfrac{1}{4}\)
\(\dfrac{1}{3}-\dfrac{1}{a\times4}=\dfrac{200}{609}\)
\(\dfrac{1}{a\times4}=\dfrac{1}{3}-\dfrac{200}{609}\)
\(\dfrac{1}{a\times4}=\dfrac{1}{203}\)
\(a\times4=203\)
\(a=\dfrac{203}{4}\)
\(\dfrac{1}{3\times7}\)+\(\dfrac{1}{7\times11}\)+\(\dfrac{1}{11\times15}\)+...+\(\dfrac{1}{a\times\left(a+4\right)}\) = \(\dfrac{50}{609}\)
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
Giải:
Ta có:
\(\left\{{}\begin{matrix}\dfrac{a}{b}=\dfrac{14}{22}\\\dfrac{c}{d}=\dfrac{11}{13}\\\dfrac{e}{f}=\dfrac{13}{17}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{b}=\dfrac{7}{11}\\\dfrac{c}{d}=\dfrac{11}{13}\\\dfrac{e}{f}=\dfrac{13}{17}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{7}=\dfrac{b}{11}\\\dfrac{c}{11}=\dfrac{d}{13}\\\dfrac{e}{13}=\dfrac{f}{17}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{7}=\dfrac{b}{11}=\dfrac{a+b}{7+11}=\dfrac{M}{18}\left(1\right)\\\dfrac{c}{11}=\dfrac{d}{13}=\dfrac{c+d}{11+13}=\dfrac{M}{24}\left(2\right)\\\dfrac{e}{13}=\dfrac{f}{17}=\dfrac{e+f}{13+17}=\dfrac{M}{30}\left(3\right)\end{matrix}\right.\)
Kết hợp \(\left(1\right);\left(2\right)\) và \(\left(3\right)\)
\(\Rightarrow M\in BC\left(18;24;30\right)\)
Mặt khác \(M\) là số tự nhiên nhỏ nhất có 4 chữ số
Nên \(M=1080\)
Vậy \(M=1080\)
Ta có : \(\frac{a}{b}=\frac{14}{22}\Rightarrow\frac{a}{14}=\frac{b}{22}=\frac{a+b}{14+22}=\frac{M}{36}\)
\(\frac{c}{d}=\frac{11}{13}\Rightarrow\frac{c}{11}=\frac{d}{13}=\frac{c+d}{11+13}=\frac{M}{24}\)
\(\frac{e}{f}=\frac{13}{17}\Rightarrow\frac{e}{13}=\frac{f}{17}=\frac{e+f}{13+17}=\frac{M}{30}\)
Nhận thấy M chia hết cho 36,24,30 => \(M⋮36,M⋮24,M⋮30\)
=> \(M\in BC\left(36,24,30\right)\)
Ta có : 36 = 22 . 32
24 = 23 . 3
30 = 2.3.5
=> \(BCNN\left(36,24,30\right)=2^3\cdot3^2\cdot5=360\)
=> \(BC\left(36,24,30\right)=B\left(360\right)=\left\{0;360;720;1080\right\}\)
Vậy số tự nhiên của M là 1080
\(\frac{a}{b}\times4+\frac{1}{6}=\frac{17}{6}\)
\(\frac{a}{b}\times4=\frac{17}{6}-\frac{1}{6}\)
\(\frac{a}{b}\times4=\frac{8}{3}\)
\(\frac{a}{b}=\frac{8}{3}\div4\)
\(\frac{a}{b}=\frac{2}{3}\)
K mk nha
Mk cảm ơn bạn nhiều
Thank you so much!
a) Số nghịch đảo của \(\frac{a}{b}\) là \(\frac{b}{a}\)
b) \(-\frac{17}{7}.x=\frac{7}{-17}\Leftrightarrow x=\frac{7}{-17}:-\frac{17}{7}=\frac{49}{289}\)