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a: \(-\dfrac{6}{13}=-\dfrac{12}{26}=\dfrac{-18}{39}=-\dfrac{24}{52}=\dfrac{-30}{65}=\dfrac{-36}{78}=\dfrac{-42}{91}\)
b: \(\dfrac{15}{-7}=\dfrac{-15}{7}=\dfrac{-30}{14}=\dfrac{-45}{21}=\dfrac{-60}{28}=\dfrac{-75}{35}=-\dfrac{90}{42}\)
e,\(3\frac{2}{7}x-\frac{1}{8}=2\frac{3}{4}\)
\(=>\frac{23}{7}x-\frac{1}{8}=\frac{11}{4}\)
\(=>\frac{23}{7}x=\frac{11}{4}+\frac{1}{8}=\frac{23}{8}\)
\(=>x=\frac{23}{8}:\frac{23}{7}\)
\(=>x=\frac{7}{8}\)
b) \(5\frac{1}{4}.\frac{3}{8}+10\frac{3}{4}.\frac{3}{8}\)
\(=\left(5\frac{1}{4}+10\frac{3}{4}\right).\frac{3}{8}\)
\(=16.\frac{3}{8}=6\)
c) \(6\frac{1}{5}.\frac{-2}{7}+14\frac{4}{5}.\frac{-2}{7}\)
\(=\left(6\frac{1}{5}+14\frac{4}{5}\right).\frac{-2}{7}\)
\(=21.\frac{-2}{7}=-6\)
a) Ư(60):{ 1;2;3;4;5;6;10;12;15;20;30;60}
Ư(84):{ 1;2;4;6;7;12;14;21;42;84}
Ư(120):{ 1;2;3;4;5;6;8;10;12;15;20;24;30;40;60;120}
ƯC(60;84;120):{ 2;4;6;12}
nhưng vì x_> 6 nên x = 2,4,6
\(\dfrac{2n-1}{n+2}=\dfrac{2\left(n+2\right)}{n+2}-\dfrac{5}{n+2}=2-\dfrac{5}{n+2}\)
Để \(\dfrac{5}{n+2}\) ∈ Z
⇒ \(\left(n+2\right)\) ∈ \(Ư\left(5\right)=\left(1;-1;5;-5\right)\)
⇒ \(n\) ∈ \(\left(-1;-3;3;-7\right)\)