Cho a , b thuộc Z và b>0 . So sánh a/b và a+1/b+1
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\(a,\dfrac{a}{b}>1\Leftrightarrow a>1\cdot b=b\\ \dfrac{a}{b}< 1\Leftrightarrow a< 1\cdot b=b\\ b,\dfrac{a}{b}=\dfrac{a\left(b+1\right)}{b\left(b+1\right)}=\dfrac{ab+a}{b^2+b}\\ \dfrac{a+1}{b+1}=\dfrac{b\left(a+1\right)}{b\left(b+1\right)}=\dfrac{ab+b}{b^2+b}\\ \forall a=b\Leftrightarrow\dfrac{a}{b}=\dfrac{a+1}{b+1}\\ \forall a>b\Leftrightarrow\dfrac{a}{b}>\dfrac{a+1}{b+1}\\ \forall a< b\Leftrightarrow\dfrac{a}{b}< \dfrac{a+1}{b+1}\)
\(c,\forall a>b\Leftrightarrow\dfrac{a}{b}-1=\dfrac{a-b}{b}>\dfrac{a-b}{b+n}\left(b< b+n;a-b>0\right)=\dfrac{a+n}{b+n}-1\\ \Leftrightarrow\dfrac{a}{b}>\dfrac{a+n}{b+n}\\ \forall a< b\Leftrightarrow1-\dfrac{a}{b}=\dfrac{b-a}{b}>\dfrac{b-a}{b+n}\left(b< b+n;b-a>0\right)=1-\dfrac{a+n}{b+n}\\ \Leftrightarrow1-\dfrac{a}{b}>1-\dfrac{a+n}{b+n}\Leftrightarrow\dfrac{a}{b}>\dfrac{a+n}{b+n}\\ \forall a=b\Leftrightarrow\dfrac{a+n}{b+n}=\dfrac{a}{b}\left(=1\right)\)
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\(\dfrac{a}{b}=\dfrac{a\left(b+2021\right)}{b\left(b+2021\right)}=\dfrac{ab+2021a}{b\left(b+2021\right)}\\ \dfrac{a+2021}{b+2021}=\dfrac{ab+2021b}{b\left(b+2021\right)}\)
Vì \(b>0\Rightarrow b\left(b+2021\right)>0\)
Nếu \(a< b\Leftrightarrow\dfrac{a}{b}< \dfrac{a+2021}{b+2021}\)
Nếu \(a=b\Leftrightarrow\dfrac{a}{b}=\dfrac{a+2021}{b+2021}=1\)
Nếu \(a>b\Leftrightarrow\dfrac{a}{b}>\dfrac{a+2021}{b+2021}\)
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Ta có : \(\frac{a}{b}< \frac{a+n}{b+n}\Leftrightarrow a(b+n)< b(a+n)\)
\(\Leftrightarrow ab+an< ab+bn\Leftrightarrow a< b\)vì n > 0
Như vậy : \(\frac{a}{b}< \frac{a+n}{b+n}\Leftrightarrow a< b\)
Ta lại có : \(\frac{a}{b}>\frac{a+n}{b+n}\Leftrightarrow a(b+n)>b(a+n)\)
\(\Leftrightarrow ab+an>ab+bn\Leftrightarrow an>bn\Leftrightarrow a>b\)
Như vậy : \(\frac{a}{b}>\frac{a+n}{b+n}\Leftrightarrow a>b\)
\(\frac{a}{b}