- So sánh cặp số hữu tỉ sau : \(\frac{1001}{1000}\)và\(\frac{1002}{1003}\)
2. a,Cho số hữu tỉ x =\(\frac{a-3}{2a}\).Với giá trị nào của a thì x kaf số nguyên ?
b, CMR : Nếu\(\frac{a+2009}{a-2009}\)=\(\frac{b+2010}{b-2010}\)thì\(\frac{a}{2009}\)=\(\frac{b}{2010}\).
1.\(\frac{1001}{1000}>\frac{1000}{1000}=1=\frac{1003}{1003}>\frac{1002}{1003}\Rightarrow\frac{1001}{1000}>\frac{1002}{1003}\)
2.a) \(x=\frac{a-3}{2a}\left(a\ne0\right)\)
\(=\frac{1}{2}\left(1-\frac{3}{a}\right)\inℤ\)
\(\Leftrightarrow\hept{\begin{cases}1-\frac{3}{a}\inℤ\\1-\frac{3}{a}⋮2\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\frac{3}{a}\inℤ\\\frac{3}{a}\equiv1\left(mod2\right)\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\\\frac{3}{a}\equiv1\left(mod2\right)\end{cases}}\)
Ta có bảng :
Vậy \(a\in\left\{\pm1;\pm3\right\}\)
b)Ta có:\(\frac{a+2009}{a-2009}=1+\frac{4018}{a-2009}\left(a\ne2009\right)\)
\(\frac{b+2010}{b-2010}=1+\frac{4020}{b-2010}\left(b\ne2010\right)\)
\(\Rightarrow\frac{4018}{a-2009}=\frac{4020}{b-2010}\)
\(\Rightarrow\frac{a-2009}{4018}=\frac{b-2010}{4020}\)
\(\Rightarrow\frac{a-2009}{2009}=\frac{b-2010}{2010}\)
\(\Rightarrow\frac{a}{2009}-1=\frac{b}{2010}-1\)
\(\Rightarrow\frac{a}{2009}=\frac{b}{2010}\)
Thanks!