\(a,A=\dfrac{1}{7.9}\) và \(B=\dfrac{1}{7}-\dfrac{1}{9}\)
\(\Leftrightarrow2A=\dfrac{2}{7.9}=\dfrac{1}{7}-\dfrac{1}{9}\)
Vì \(2A=B\left(=\dfrac{1}{7}-\dfrac{1}{9}\right)\)
\(\Leftrightarrow A< B\)
\(b,A=\dfrac{15}{301}\) và \(B=\dfrac{25}{499}\)
Ta thấy: \(\dfrac{15}{301}< \dfrac{15}{300}\) và \(\dfrac{25}{499}>\dfrac{25}{500}\)
\(\dfrac{15}{300}=\dfrac{1}{20}\) và \(\dfrac{25}{500}=\dfrac{1}{20}\)
Vì \(\dfrac{1}{20}=\dfrac{1}{20}\Leftrightarrow\dfrac{15}{300}=\dfrac{25}{500}\)
\(\Leftrightarrow\dfrac{15}{301}< \dfrac{1}{20}< \dfrac{25}{499}\)
\(\Leftrightarrow\dfrac{15}{301}< \dfrac{25}{499}\)
\(\Leftrightarrow A< B\)
\(c,A=\dfrac{5.6}{9.25}\) và \(B=\dfrac{18.4-18}{8.9+7.9}\)
\(A=\dfrac{5.6}{9.25}=\dfrac{1.2}{3.5}=\dfrac{2}{15}\)
\(B=\dfrac{18.4-18}{8.9+7.9}=\dfrac{18\left(4-1\right)}{9\left(8+7\right)}=\dfrac{18.3}{9.15}=\dfrac{2.1}{1.5}=\dfrac{2}{5}\)
Vì \(\dfrac{2}{15}< \dfrac{2}{5}\)
\(\Leftrightarrow A< B\)
Chúc bạn học tốt!

Gọi \(ƯC\left(3n+2;2n+1\right)\)là \(d\)
\(\Leftrightarrow\left\{{}\begin{matrix}3n+2⋮d\\2n+1⋮d\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2\left(3n+2\right)⋮d\\3\left(2n+1\right)⋮d\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}6n+4⋮d\\6n+3⋮d\end{matrix}\right.\)
\(\Leftrightarrow6n+4-\left(6n+3\right)⋮d\)
\(\Leftrightarrow6n+4-6n-3⋮d\)
\(\Leftrightarrow1⋮d\)
\(\Leftrightarrow d=\pm1\)
Vậy \(\dfrac{3n+2}{2n+1}\) là phân số tổi giản \(\forall\) \(n\in Z\)
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\(\left|x^2-5\right|+\left|5-x^2\right|=40\)
\(\Leftrightarrow\left|x^2-5\right|+\left|x^2-5\right|=40\)
\(\Leftrightarrow2.\left|x^2-5\right|=40\)
\(\Leftrightarrow\left|x^2-5\right|=20\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-5=20\\x^2-5=-20\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=25\\x^2=-15\end{matrix}\right.\)
Vì \(x^2\) luôn là số nguyên dương.
\(\Leftrightarrow x^2=25\)
\(\Leftrightarrow x=\pm5\)
Vậy \(x=\pm5\) thì \(\left|x^2-5\right|+\left|5-x^2\right|=40\)
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\(x\left(x-3\right)< 0\)
Vì \(x>x-3\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>0\\x-3< 0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x>0\\x< 3\end{matrix}\right.\)\(\Leftrightarrow x\in\left\{1;2\right\}\)
Vậy \(x\in\left\{1;2\right\}\) thì \(x\left(x-3\right)< 0\)