Question

[Lý Phương Chi]

1. 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\(\frac{a}{2009}\)=\(\frac{b}{2010}\).

Answer 1

[Yuri Sweet[𝕿𝖊𝖆𝖒 𝕹𝖊𝖕𝖆𝖑]]

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 :

\(a\)	\(1\)	\(-1\)	\(3\)	\(-3\)
\(\frac{3}{a}\)	\(3\)	\(-3\)	\(1\)	\(-1\)
\(1-\frac{3}{a}\)	\(-2\)	\(4\)	\(0\)	\(2\)
\(x\)	\(-1\)	\(2\)	\(0\)	\(1\)

Vậy \(a\in\left\{\pm1;\pm3\right\}\)

b)Ta có:\(\frac{a+2009}{a-2009}=1+\frac{4018}{a-2009}\left(a\ne2009\right)\)

\(\frac{b+2010}{b-2010}=1+\frac{4020}{b-2010}\left(b\ne2010\right)\)

\(\Rightarrow\frac{4018}{a-2009}=\frac{4020}{b-2010}\)

\(\Rightarrow\frac{a-2009}{4018}=\frac{b-2010}{4020}\)

\(\Rightarrow\frac{a-2009}{2009}=\frac{b-2010}{2010}\)

\(\Rightarrow\frac{a}{2009}-1=\frac{b}{2010}-1\)

\(\Rightarrow\frac{a}{2009}=\frac{b}{2010}\)

Answer 2

[Lý Phương Chi]

Thanks!