Bài 22:
\(\dfrac{n+1}{2n+7}-\dfrac{n+5}{2n+5}\)
\(=\dfrac{\left(n+1\right)\left(2n+5\right)-\left(n+5\right)\left(2n+7\right)}{\left(2n+7\right)\left(2n+5\right)}\)
\(=\dfrac{2n^2+7n+5-2n^2-7n-10n-35}{\left(2n+7\right)\left(2n+5\right)}\)
\(=\dfrac{-10n-30}{\left(2n+7\right)\left(2n+5\right)}=\dfrac{-10\left(n+3\right)}{\left(2n+7\right)\left(2n+5\right)}< 0\)
=>\(\dfrac{n+1}{2n+7}< \dfrac{n+5}{2n+5}\)
Bài 20:
a: Để B là số nguyên thì \(2m+3⋮m+1\)
=>\(2m+2+1⋮m+1\)
=>\(1⋮m+1\)
=>\(m+1\in\left\{1;-1\right\}\)
=>\(m\in\left\{0;-2\right\}\)
b: Gọi d=ƯCLN(2m+3;m+1)
=>\(\left\{{}\begin{matrix}2m+3⋮d\\m+1⋮d\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2m+3⋮d\\2m+2⋮d\end{matrix}\right.\)
=>\(2m+3-2m-2⋮d\)
=>\(1⋮d\)
=>d=1
=>ƯCLN(2m+3;m+1)=1
=>\(\dfrac{2m+3}{m+1}\) là phân số tối giản
help bài 3 câu a ,b, và c help
