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\(a,\dfrac{a}{b}>1\Leftrightarrow a>1\cdot b=b\\ \dfrac{a}{b}< 1\Leftrightarrow a< 1\cdot b=b\\ b,\dfrac{a}{b}=\dfrac{a\left(b+1\right)}{b\left(b+1\right)}=\dfrac{ab+a}{b^2+b}\\ \dfrac{a+1}{b+1}=\dfrac{b\left(a+1\right)}{b\left(b+1\right)}=\dfrac{ab+b}{b^2+b}\\ \forall a=b\Leftrightarrow\dfrac{a}{b}=\dfrac{a+1}{b+1}\\ \forall a>b\Leftrightarrow\dfrac{a}{b}>\dfrac{a+1}{b+1}\\ \forall a< b\Leftrightarrow\dfrac{a}{b}< \dfrac{a+1}{b+1}\)
\(c,\forall a>b\Leftrightarrow\dfrac{a}{b}-1=\dfrac{a-b}{b}>\dfrac{a-b}{b+n}\left(b< b+n;a-b>0\right)=\dfrac{a+n}{b+n}-1\\ \Leftrightarrow\dfrac{a}{b}>\dfrac{a+n}{b+n}\\ \forall a< b\Leftrightarrow1-\dfrac{a}{b}=\dfrac{b-a}{b}>\dfrac{b-a}{b+n}\left(b< b+n;b-a>0\right)=1-\dfrac{a+n}{b+n}\\ \Leftrightarrow1-\dfrac{a}{b}>1-\dfrac{a+n}{b+n}\Leftrightarrow\dfrac{a}{b}>\dfrac{a+n}{b+n}\\ \forall a=b\Leftrightarrow\dfrac{a+n}{b+n}=\dfrac{a}{b}\left(=1\right)\)
Cho \(a,b\in\mathbb{Z},b>0\). So sánh hai số hữu tỉ \(\dfrac{a}{b}\) và \(\dfrac{a+2001}{b+2001}\) ?
Xét tích \(a\left(b+2001\right)=ab+2001a\).
\(b\left(a+2001\right)=ab+2001b\). Vì \(b>0\) nên \(b+2001>0\).
a) Nếu \(a>b\) thì \(ab+2001a>ab+2001b\)
\(a\left(b+2001\right)>b\left(a+2001\right)\)
\(\Rightarrow\dfrac{a}{b}>\dfrac{a+2001}{b+2001}\) (theo bài 5).
b) Tương tự (theo bài 5) nếu \(a< b\) thì \(\Rightarrow\dfrac{a}{b}< \dfrac{a+2001}{b+2001}\).
c) Nếu \(a=b\) thì rõ ràng \(\dfrac{a}{b}=\dfrac{a+2001}{b+2001}\).
Nếu a,b cùng dấu thì \(\dfrac{a}{b}\ge0\)
Nếu a,b khác dấu thì \(\dfrac{a}{b}< 0\)
\(\left[{}\begin{matrix}a\ge0,b>0\\a\le0,b< 0\end{matrix}\right.\Rightarrow\dfrac{a}{b}\ge0\\ \left[{}\begin{matrix}a\ge0,b< 0\\a\le0,b>0\end{matrix}\right.\Rightarrow\dfrac{a}{b}\le0\)
Ta có :
\(B=\dfrac{2009^{2010}-2}{2009^{2011}-2}< 1\)
\(\Leftrightarrow B< \dfrac{2009^{2010}-2+2011}{2009^{2011}-2+2011}=\dfrac{2009^{2010}+2009}{2009^{2011}+2009}=\dfrac{2009\left(2009^{2009}+1\right)}{2009\left(2009^{2010}+1\right)}=\dfrac{2009^{2009}+1}{2009^{2010}+1}=A\)
\(\Leftrightarrow A>B\)
Ta có
\(\dfrac{a}{b}=\dfrac{a\left(b+2010\right)}{b\left(b+2010\right)}=\dfrac{ab+2010a}{b\left(b+2010\right)}\) (1)
\(\dfrac{a+2010}{b+2010}=\dfrac{b\left(a+2010\right)}{b\left(b+2010\right)}=\dfrac{ab+2010b}{b\left(2010+b\right)}\) (2)
Nếu \(a=b\Rightarrow2010a=2010b\) nên từ 1 và 2 suy ra \(\dfrac{a}{b}=\dfrac{a+2010}{b+2010}\)
Nếu a>b \(\Rightarrow2010a>2010b\) nên tư 1 và 2 suy ra \(\dfrac{a}{b}>\dfrac{a+2010}{b+2010}\)
Nếu a<b \(\Rightarrow2010a< 2010b\) nên từ 1 va 2 suy ra \(\dfrac{a}{b}< \dfrac{a+2010}{b+2010}\)