Cho a ,b \(\in Z\) , b > 0 , So sánh hai số hữu tỉ \(\dfrac{a}{b}\) và \(\dfrac{a+2001}{b+2001}\) .
Giải
Ta có:
Nếu: \(a>b\)
\(\Rightarrow\dfrac{a}{b}>1\Rightarrow\dfrac{a+2011}{b+2011}>1\Rightarrow\dfrac{a}{b}>\dfrac{a+2011}{b+2011}\)
Nếu: \(a< b\)
\(\Rightarrow\dfrac{a}{b}< 1\Rightarrow\dfrac{a+2011}{b+2011}< 1\Rightarrow\dfrac{a}{b}< \dfrac{a+2011}{b+2011}\)
Nếu \(a=b\)
\(\Rightarrow\dfrac{a}{b}=1\Rightarrow\dfrac{a+2011}{b+2011}=1\Rightarrow\dfrac{a}{b}=\dfrac{a+2011}{b+2011}=1\)
Vậy...
Xét tích :
\(a\left(b+2001\right)=ab+a2001\)
\(b\left(a+2001\right)=ab+b2001\)
Vì \(b>0\Leftrightarrow b+2001>0\)
+) Nếu \(a>b\Leftrightarrow ab+2001a>ab+2001b\)
\(\Leftrightarrow a\left(b+2001\right)>b\left(a+2001\right)\)
\(\Leftrightarrow\dfrac{a}{b}>\dfrac{a+2001}{b+2001}\)
+) Tương tự
Nếu \(a< b\Leftrightarrow\dfrac{a}{b}< \dfrac{a+2001}{b+2001}\)
+) Nếu \(a=b\Leftrightarrow\dfrac{a}{b}=\dfrac{a+2001}{b+2001}\)
Xét tích :
a( b + 2001 ) = ab + 2001a .
b(a+ 2001 ) = ab + 2001b . Vì b > 0 nen b + 2001 > 0 .
a) Nếu a > b thì ab + 2001 a > ab + 2001b .
a(b + 2001 ) < b ( a + 2001 )
\(\Rightarrow\dfrac{a}{b}>\dfrac{a+2001}{b+2001}\)
b) tương tự nếu a < b thì \(\Rightarrow\dfrac{a}{b}< \dfrac{a+2001}{b+2001}\)
c) Nếu a = b thì rõ ràng \(\dfrac{a}{b}=\dfrac{a+2001}{b+2001}\)