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= 54/6. [6/1.3.7 +6/3.7.9 +6/7.9.13 + 6/9.13.15]
=54/6 [1/1.3-1/3.7 + 1/3.7 - 1/7.9 + 1/7.9 - 1/9.13 + 1/9.13 -1/13.15]
=54/6[1/3- 1/13.15]
=54/6 .64/195
=192/65
\(\left(\frac{1}{6}+\frac{1}{10}+\frac{1}{15}\right):\left(\frac{1}{6}+\frac{1}{10}-\frac{1}{15}\right)\)
=\(\frac{1}{3}+\frac{1}{5}\)
=\(\frac{5}{3}\)
\(\frac{a}{b}.4+\frac{1}{6}=\frac{17}{6}\)
\(\frac{a}{b}.4\) \(=\frac{17}{6}-\frac{1}{6}\)
\(\frac{a}{b}.4\) \(=\frac{8}{3}\)
\(\frac{a}{b}=\frac{8}{3}:4\)
\(\frac{a}{b}=\frac{2}{3}\)
\(\frac{a}{b}.4=\frac{17}{6}-\frac{1}{6}\)
\(\frac{a}{b}.4=\frac{16}{6}\)
\(\frac{a}{b}=\frac{16}{6}:4\)
\(\frac{a}{b}=\frac{16}{6}.\frac{1}{4}=\frac{16}{6.4}=\frac{2}{3}\)
Ta có: \(\frac{y}{3}+y\cdot\frac{2}{3}=\frac{y}{12}+\frac{11}{6}\)
=> \(\frac{y}{3}+\frac{2y}{3}=\frac{y}{12}+\frac{22}{12}\)
=> \(\frac{y+2y}{3}=\frac{y+22}{12}\)
=> \(\frac{3y}{3}=\frac{y+22}{12}\)
=> \(y=\frac{y+22}{12}\)
=> y + 22 = 12y
=> y - 12y = 22
=> -11y = 22
=> y = 22 : (-11) = -2
Vậy y = -2
\(\frac{6}{1\times3\times7}+\frac{6}{3\times7\times9}+\frac{6}{7\times9\times13}+\frac{6}{9\times13\times15}+\frac{6}{13\times15\times9}\)
\(=\frac{1}{1\times3}-\frac{1}{3\times7}+\frac{1}{3\times7}-\frac{1}{7\times9}+\frac{1}{7\times9}-\frac{1}{9\times13}+\frac{1}{9\times13}-\frac{1}{13\times15}+\frac{1}{13\times15}-\frac{1}{15\times9}\)
\(=\frac{1}{1\times3}-\frac{1}{15\times9}=\frac{1}{3}-\frac{1}{135}=\frac{45}{135}-\frac{1}{135}=\frac{44}{135}\)