(Sửa \(cn-bm\rightarrow cn-dm\))

Ta có :

\(\left\{{}\begin{matrix}ad-bc=1\\cn-dm=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}ad=1+bc\\cn=1+dm\end{matrix}\right.\)

\(\dfrac{x}{y}=\dfrac{a}{b}.\dfrac{d}{c}=\dfrac{ad}{bc}=\dfrac{1+bc}{bc}=1+\dfrac{1}{bc}>1\left(bc>0\right)\)

\(\Rightarrow x=\dfrac{a}{b}>y=\dfrac{c}{d}\left(2\right)\)

\(\dfrac{y}{z}=\dfrac{c}{d}.\dfrac{n}{m}=\dfrac{cn}{dm}=\dfrac{1+dm}{dm}=1+\dfrac{1}{dm}>1\left(dc>0\right)\)

\(\Rightarrow y=\dfrac{c}{d}>z=\dfrac{m}{n}\left(2\right)\)

\(\left(1\right);\left(2\right)\Rightarrow x>y>z\)

a: \(x_1=x_2=x_3=x_4\)

Bài 1:

1)mik ko biết trục số ở đâu nên tham khảo:

2

-0,75 <5/3

\(a,\dfrac{a}{b}>1\Leftrightarrow a>1\cdot b=b\\ \dfrac{a}{b}< 1\Leftrightarrow a< 1\cdot b=b\\ b,\dfrac{a}{b}=\dfrac{a\left(b+1\right)}{b\left(b+1\right)}=\dfrac{ab+a}{b^2+b}\\ \dfrac{a+1}{b+1}=\dfrac{b\left(a+1\right)}{b\left(b+1\right)}=\dfrac{ab+b}{b^2+b}\\ \forall a=b\Leftrightarrow\dfrac{a}{b}=\dfrac{a+1}{b+1}\\ \forall a>b\Leftrightarrow\dfrac{a}{b}>\dfrac{a+1}{b+1}\\ \forall a< b\Leftrightarrow\dfrac{a}{b}< \dfrac{a+1}{b+1}\)

\(c,\forall a>b\Leftrightarrow\dfrac{a}{b}-1=\dfrac{a-b}{b}>\dfrac{a-b}{b+n}\left(b< b+n;a-b>0\right)=\dfrac{a+n}{b+n}-1\\ \Leftrightarrow\dfrac{a}{b}>\dfrac{a+n}{b+n}\\ \forall a< b\Leftrightarrow1-\dfrac{a}{b}=\dfrac{b-a}{b}>\dfrac{b-a}{b+n}\left(b< b+n;b-a>0\right)=1-\dfrac{a+n}{b+n}\\ \Leftrightarrow1-\dfrac{a}{b}>1-\dfrac{a+n}{b+n}\Leftrightarrow\dfrac{a}{b}>\dfrac{a+n}{b+n}\\ \forall a=b\Leftrightarrow\dfrac{a+n}{b+n}=\dfrac{a}{b}\left(=1\right)\)

bằng nhau                               

Ta có: \(\frac{a}{b}=\frac{a.\left(b+1\right)}{b.\left(b+1\right)}=\frac{ab+a}{b.\left(b+1\right)}\)

          \(\frac{a+1}{b+1}=\frac{b.\left(a+1\right)}{b.\left(b+1\right)}=\frac{ab+b}{b.\left(b+1\right)}\)

Xét a>b

=>\(\frac{ab+a}{b.\left(b+1\right)}>\frac{ab+b}{b.\left(b+1\right)}\)

=>\(\frac{a}{b}>\frac{a+1}{b+1}\)

Xét a<b

=>\(\frac{ab+a}{b.\left(b+1\right)}

xảy ra 3 trường hợp:

1)a/b>c

2)a/b=c

3)a/b<c

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

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

2.Gọi 3 p/số đó là x;y;z

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

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

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

Nếu a,b cùng dấu thì \(\dfrac{a}{b}\ge0\)

Nếu a,b khác dấu thì \(\dfrac{a}{b}< 0\)

\(\left[{}\begin{matrix}a\ge0,b>0\\a\le0,b< 0\end{matrix}\right.\Rightarrow\dfrac{a}{b}\ge0\\ \left[{}\begin{matrix}a\ge0,b< 0\\a\le0,b>0\end{matrix}\right.\Rightarrow\dfrac{a}{b}\le0\)

18/31 giữ nguyên . 181818/313131=18 nhân 10101/31 nhân 10101 = 18/31

18/31=181818/313131