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Có \(\frac{a}{b}< \frac{c}{d}=>a.d< c.b\)
<=>2018a.d<2018c.b
<=>2018a.d+c.d<2018c.b+c.d
<=>d(2018a+c)<c(2018b+d)
<=>đpcm
Vì \(\frac{a}{b}< \frac{c}{d}\)
⇒ \(ad< bc\)
⇒ \(2018ad< 2018bc\)
⇒ \(2018ad+cd< 2018bc+cd\)
⇒ \(\left(2018a+c\right)d< \left(2018b+d\right)c\)
⇒ \(\frac{2018a+c}{2018b+d}< \frac{c}{d}\)
Vậy \(\frac{2018a+c}{2018b+d}< \frac{c}{d}\) (ĐPCM)
Vì a/b < c/d (Với a,b,c,d thuộc N*)
=> ad<bc
=> 2018ad < 2018bc
=> 2018ad + cd < 2018bc +cd
=> (2018a + c).d < (2018b+d).c
=> 2018a +c / 2018b + d < c/d
Ta có:a/b<c/d<=>a.d<b.c
<=>2018a.d<2018b.c
<=>2018a.d+c.d<2018b.c+d.c
<=>d(2018a+c)<c(2018b+d)
<=>2018a+c/2018b+d<c/d(dpcm)
Ta có: Để \(\frac{2018\cdot a+c}{2018\cdot b+d}< \frac{c}{d}\Rightarrow\left(2018\cdot a+c\right)\cdot d< \left(2018\cdot b+d\right)\cdot c\)
\(2018\cdot a\cdot d+c\cdot d< 2018\cdot b\cdot c+c\cdot d\)
\(2018\cdot a\cdot d< 2018\cdot b\cdot c\)(bỏ cả 2 vế đi \(c\cdot d\))(gọi là (1))
Vì \(\frac{a}{b}< \frac{c}{d}\Rightarrow a\cdot d< b\cdot c\Rightarrow2018\cdot a\cdot d< 2018\cdot b\cdot c=\left(1\right)\)Mà (1) bằng \(\frac{2018\cdot a+c}{2018\cdot b+d}< \frac{c}{d}\) (điều phải chứng minh)
2.
\(B=\frac{2018}{1}+\frac{2017}{2}+...+\frac{1}{2018}\)
\(=\left(\frac{1}{2018}+1\right)+\left(\frac{2}{2017}+1\right)+...+\left(\frac{2017}{2}+1\right)+1\)
\(=\frac{2018+1}{2018}+\frac{2017+2}{2017}+...+\frac{2+2017}{2}+1\)
\(=\frac{2019}{2019}+\frac{2019}{2018}+...+\frac{2019}{2}\)
\(=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}\right)=2019A\)
\(\Rightarrow\frac{A}{B}=\frac{A}{2019A}=\frac{1}{2019}\)
