Áp dụng BĐT Cauchy–Schwarz ta được:

\(x=\dfrac{2017}{\sqrt{2018}}+\dfrac{2018}{\sqrt{2017}}\ge\dfrac{\left(\sqrt{2018}+\sqrt{2017}\right)^2}{\sqrt{2018}+\sqrt{2017}}=\sqrt{2018}+\sqrt{2017}=y\)

Dấu \("="\Leftrightarrow\dfrac{2017}{\sqrt{2018}}=\dfrac{2018}{\sqrt{2017}}\Leftrightarrow2017=2018\left(vô.lí\right)\)

Vậy đẳng thức ko xảy ra hay \(x>y\)

\(-\frac{13}{15}+-\frac{2}{15}=-1;-\frac{14}{16}+-\frac{2}{16}\)

Vì \(-\frac{2}{15}< -\frac{2}{16}\Rightarrow\frac{-13}{15}< -\frac{14}{16}\)

2.Gọi 3 p/số đó là x;y;z

\(-\frac{5}{8}< x< y< z< -\frac{3}{5}\)

\(-\frac{100}{160}< x< y< z< -\frac{96}{160}\)

\(\Rightarrow x=-\frac{99}{160};y=-\frac{98}{160}=-\frac{49}{80};z=-\frac{97}{160}\)

a: \(0.2=\dfrac{2}{10}\)

10>7

=>\(\dfrac{2}{10}< \dfrac{2}{7}\)

=>\(\dfrac{2}{7}>0.2\)

b: \(-\dfrac{1^5}{6}=\dfrac{-1}{6}=\dfrac{-3}{18}\)

\(\dfrac{8}{-9}=-\dfrac{16}{18}\)

mà -3>-16

nên \(-\dfrac{1^5}{6}>\dfrac{8}{-9}\)

c: \(\dfrac{2017}{2016}>1\)

\(1>\dfrac{2017}{2018}\)

Do đó: \(\dfrac{2017}{2016}>\dfrac{2017}{2018}\)

d: \(-\dfrac{249}{333}=\dfrac{-249:3}{333:3}=\dfrac{-83}{111}\)

e: \(\dfrac{5^1}{3}=\dfrac{5}{3}=\dfrac{15}{9}\)

\(\dfrac{4^8}{9}=\dfrac{65536}{9}\)

mà 15<65536

nên \(\dfrac{5^1}{3}< \dfrac{4^8}{9}\)

f: 13,589<13,612

a)-17/23=-171717/232323

b)-265/317<-83/111

c)2002/2003<14/13

d)-27/463<1/3

y = 0 , 75 = 3 4 = 15 20   n ê n   x > y

97/98 < 2018/2019

-25/26 < -18/31

-9/11 > -13/12

1.

Ta có:

\(-\dfrac{13}{15}=-1+\dfrac{2}{15}\)

\(-\dfrac{14}{16}=-1+\dfrac{2}{16}\)

Do đó:\(-\dfrac{14}{16}< -\dfrac{13}{15}\)

2.

Ta có:\(-\dfrac{3}{5}>x>-\dfrac{5}{8}\)

=>\(-\dfrac{96}{160}>x>-\dfrac{100}{160}\)

=>3 phân số cần tìm là \(-\dfrac{97}{160};-\dfrac{98}{160};-\dfrac{99}{160}\)