Cho A / B = 3 / 5, B / C = 7 / 11, C / D = 2 / 3 và A + B + C + D == 1161. Tìm giá trị của B.
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a, \(\dfrac{4}{7}\). \(\dfrac{a}{b}\) - \(\dfrac{1}{3}\) = \(\dfrac{1}{21}\)
\(\dfrac{4}{7}\).\(\dfrac{a}{b}\) = \(\dfrac{1}{21}\) + \(\dfrac{1}{3}\)
\(\dfrac{4}{7}\).\(\dfrac{a}{b}\) = \(\dfrac{8}{21}\)
\(\dfrac{a}{b}\) = \(\dfrac{8}{21}\):\(\dfrac{4}{7}\)
\(\dfrac{a}{b}\) = \(\dfrac{2}{3}\)
b, \(\dfrac{a}{b}\) + \(\dfrac{2}{3}\).\(\dfrac{1}{3}\) = \(\dfrac{2}{3}\)
\(\dfrac{a}{b}\) + \(\dfrac{2}{9}\) = \(\dfrac{2}{3}\)
\(\dfrac{a}{b}\) = \(\dfrac{2}{3}\) - \(\dfrac{2}{9}\)
\(\dfrac{a}{b}\) = \(\dfrac{4}{9}\)
c, \(\dfrac{a}{b}\) - \(\dfrac{1}{2}.\)\(\dfrac{2}{3}\) = \(\dfrac{2}{7}\)
\(\dfrac{a}{b}\) - \(\dfrac{1}{3}\) = \(\dfrac{2}{7}\)
\(\dfrac{a}{b}\) = \(\dfrac{2}{7}\) + \(\dfrac{1}{3}\)
\(\dfrac{a}{b}\) = \(\dfrac{13}{21}\)
d, \(\dfrac{11}{13}\): \(\dfrac{a}{b}\): \(\dfrac{2}{3}\) = 2\(\dfrac{7}{13}\)
\(\dfrac{11}{13}\): \(\dfrac{a}{b}\):\(\dfrac{2}{3}\) = \(\dfrac{33}{13}\)
\(\dfrac{11}{13}\): \(\dfrac{a}{b}\) = \(\dfrac{33}{13}\) \(\times\) \(\dfrac{2}{3}\)
\(\dfrac{11}{13}\): \(\dfrac{a}{b}\) = \(\dfrac{66}{39}\)
\(\dfrac{a}{b}\) = \(\dfrac{11}{13}\) : \(\dfrac{66}{39}\)
\(\dfrac{a}{b}\) = \(\dfrac{1}{2}\)
Lời giải:
$\frac{A}{B}=\frac{3}{5}\Rightarrow A=\frac{3}{5}B$
$\frac{B}{C}=\frac{7}{11}\Rightarrow C=\frac{11}{7}B$
$\frac{C}{D}=\frac{2}{3}\Rightarrow D=\frac{3}{2}C=\frac{3}{2}.\frac{11}{7}B=\frac{33}{14}B$
$A+B+C+D=1161$
$\frac{3}{5}B+B+\frac{11}{7}B+\frac{33}{14}B=1161$
$B.(\frac{3}{5}+1+\frac{11}{7}+\frac{33}{14})=1161$
$B.\frac{387}{70}=1161$
$B=210$