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I call the number of students didn't take part in Mathemas Competition in class A & B are a & b respectively
(with condition: a & b \(\in\)N*)
,From the theme, we have: \(\frac{1}{3}b+a=\frac{1}{5}a+b\)(Because rhe number of students in 2 class is the same)
\(\Leftrightarrow\frac{b}{3}-b+a-\frac{a}{5}=0\Leftrightarrow\left(-\frac{2b}{3}\right)+\frac{4a}{5}=0\)
\(\Leftrightarrow\frac{4a}{5}-\frac{2b}{3}=0\Leftrightarrow\frac{12a-10b}{15}=0\Leftrightarrow12a=10b\)
\(\Rightarrow\frac{a}{b}=\frac{10}{12}=\frac{5}{6}\)
So the ratio of the number of student didn't take part of Class A & B is \(\frac{5}{6}.\)
a, \(\dfrac{20}{x}=\dfrac{-12}{15}\Rightarrow x=\dfrac{20.15}{-12}\Rightarrow x=-25\)
\(b,\dfrac{-15}{35}=\dfrac{27}{x}\Rightarrow x=\dfrac{35.27}{-15}\Rightarrow x=-63\)
\(c,\dfrac{\dfrac{4}{5}}{1\dfrac{2}{5}}=\dfrac{2\dfrac{2}{5}}{x}\Rightarrow x=\dfrac{2\dfrac{2}{5}.1\dfrac{2}{5}}{\dfrac{4}{5}}\Rightarrow x=\dfrac{\dfrac{84}{25}}{\dfrac{4}{5}}\Rightarrow x=\dfrac{21}{5}\)
\(d,\dfrac{x}{1\dfrac{1}{4}}=\dfrac{1\dfrac{1}{5}}{2}\Rightarrow x=\dfrac{1\dfrac{1}{4}.1\dfrac{1}{5}}{2}\Rightarrow x=\dfrac{\dfrac{3}{2}}{2}\Rightarrow x=\dfrac{3}{4}\)
\(e,\dfrac{\dfrac{1}{2}}{1\dfrac{1}{4}}=\dfrac{x}{3\dfrac{1}{3}}\Rightarrow x=\dfrac{\dfrac{1}{2}.3\dfrac{1}{3}}{1\dfrac{1}{4}}\Rightarrow x=\dfrac{\dfrac{5}{3}}{\dfrac{5}{4}}\Rightarrow x=\dfrac{4}{3}\)
a.
\(\frac{2}{-7}< 0\)
\(0< 0,25\)
\(\Rightarrow\frac{2}{-7}< 0,25\)
\(\Rightarrow y< x\)
b.
\(-\frac{3}{101}< 0\)
\(0< \frac{1}{97}\)
\(\Rightarrow\frac{-3}{101}< \frac{1}{97}\)
\(\Rightarrow x< y\)
c.
\(\frac{4}{-3}< 0\)
\(0< \frac{-1}{-103}\)
\(\Rightarrow\frac{4}{-3}< \frac{-1}{-103}\)
\(\Rightarrow x< y\)