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-13/38=0,464285714
29/-88=0,329545454
=>-13/38 lớn hơn 29/-88
\(\frac{-13}{38}=\frac{-13\cdot44}{38\cdot44}=-\frac{572}{1672}.\)
\(\frac{29}{-88}=-\frac{29}{88}=\frac{-29\cdot19}{88\cdot19}=\frac{-551}{1672}\)
Vì -572 < - 551 nên \(-\frac{572}{1672}< -\frac{551}{1672}\Rightarrow-\frac{13}{38}< \frac{29}{-88}\)
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(Sửa \(cn-bm\rightarrow cn-dm\))
Ta có :
\(\left\{{}\begin{matrix}ad-bc=1\\cn-dm=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}ad=1+bc\\cn=1+dm\end{matrix}\right.\)
\(\dfrac{x}{y}=\dfrac{a}{b}.\dfrac{d}{c}=\dfrac{ad}{bc}=\dfrac{1+bc}{bc}=1+\dfrac{1}{bc}>1\left(bc>0\right)\)
\(\Rightarrow x=\dfrac{a}{b}>y=\dfrac{c}{d}\left(2\right)\)
\(\dfrac{y}{z}=\dfrac{c}{d}.\dfrac{n}{m}=\dfrac{cn}{dm}=\dfrac{1+dm}{dm}=1+\dfrac{1}{dm}>1\left(dc>0\right)\)
\(\Rightarrow y=\dfrac{c}{d}>z=\dfrac{m}{n}\left(2\right)\)
\(\left(1\right);\left(2\right)\Rightarrow x>y>z\)
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c) \(\dfrac{27}{26}\)và\(\dfrac{38}{37}\)
Ta có: \(\dfrac{27}{26}=1+\dfrac{1}{26}\); \(\dfrac{38}{37}=1+\dfrac{1}{37}\)
Vì \(\dfrac{1}{26}>\dfrac{1}{37}\) nên \(\dfrac{27}{26}>\dfrac{38}{37}\)
c: \(\dfrac{27}{26}-1=\dfrac{1}{26}\)
\(\dfrac{38}{37}-1=\dfrac{1}{37}\)
mà \(\dfrac{1}{26}>\dfrac{1}{37}\)
nên \(\dfrac{27}{26}>\dfrac{38}{37}\)
So sánh 2 tích chéo ta có:
\(\left(-13\right)\left(-88\right)=1144\)
\(29.38=1102\)
\(1144>1102\)
\(\Leftrightarrow\dfrac{-13}{38}>\dfrac{29}{-88}\)
\(\dfrac{-13}{38}=\dfrac{-572}{1672}\)
\(\dfrac{29}{-88}=\dfrac{-551}{1672}\)
Ta thấy \(-572< -551\) nên \(\dfrac{-572}{1672}< \dfrac{-551}{1672}\) do đó \(\dfrac{-13}{38}< \dfrac{29}{-88}\)