Question

[Trần Nguyệt Hà]

a, chứng minh nếu a/b <c/d thì a/b<a/c+b/d<c/d

b, áp dụng viết 5 phân số >-1/3 và <-1/4

c, so sánh 2 số hữu tỉ

A=\(\dfrac{10^{2017}+1}{10^{2016}+1}\) và B=\(\dfrac{10^{2018}+1}{10^{2017}+1}\)

Answer 1

[Nguyễn Thanh Hằng]

Bài 1 :

a, Ta có :

\(\dfrac{a}{b}< \dfrac{c}{d}\Leftrightarrow ad< bc\)

\(\Leftrightarrow ad+ab< bc+ab\)

\(\Leftrightarrow a\left(b+d\right)< b\left(a+c\right)\)

\(\Leftrightarrow\dfrac{a}{b}< \dfrac{a+c}{b+d}\) \(\left(1\right)\)

Mà \(ad< bc\)

\(\Leftrightarrow ad+cd< bc+cd\)

\(\Leftrightarrow d\left(a+c\right)< c\left(b+d\right)\)

\(\Leftrightarrow\dfrac{a+c}{b+d}< \dfrac{c}{d}\) \(\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\dfrac{a}{b}< \dfrac{a+c}{b+d}< \dfrac{c}{d}\rightarrowđpcm\)

b) \(\dfrac{-1}{3}=\dfrac{-16}{48}< \dfrac{-15}{48};\dfrac{-14}{48};\dfrac{-13}{48}< \dfrac{-12}{48}=\dfrac{-1}{4}\)

Answer 2

[Nguyễn Thanh Hằng]

Ta thấy :

\(\left\{{}\begin{matrix}A=\dfrac{10^{2017}+1}{10^{2016}+1}>1\\B=\dfrac{10^{2018}+1}{10^{2017}+1}>1\end{matrix}\right.\)

Áp dụng tính chất \(\dfrac{a}{b}>1\Leftrightarrow\dfrac{a+m}{b+m}\) ta có :

\(B=\dfrac{10^{2018}+1}{10^{2017}+1}>\dfrac{10^{2018}+1+9}{10^{2017}+1+9}=\dfrac{10^{2018}+10}{10^{2017}+10}=\dfrac{10\left(10^{2017}+1\right)}{10\left(10^{2016}+1\right)}=\dfrac{10^{2017}+1}{10^{2016}+1}=A\)

\(\Leftrightarrow B>A\)