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\(a,A=\dfrac{-3\left(2n-3\right)-8}{2n-3}=-3-\dfrac{8}{2n-3}\in Z\\ \Leftrightarrow2n-3\inƯ\left(8\right)=\left\{-8;-4;-2;-1;1;2;4;8\right\}\\ \Leftrightarrow n\in\left\{1;2\right\}\left(n\in Z\right)\)
\(b,\dfrac{ab}{a+2b}=\dfrac{3}{2}\Leftrightarrow\dfrac{a+2b}{ab}=\dfrac{2}{3}\Leftrightarrow\dfrac{1}{b}+\dfrac{2}{a}=\dfrac{2}{3}\\ \dfrac{bc}{b+2c}=\dfrac{4}{3}\Leftrightarrow\dfrac{b+2c}{bc}=\dfrac{3}{4}\Leftrightarrow\dfrac{1}{c}+\dfrac{2}{b}=\dfrac{3}{4}\\ \dfrac{ca}{c+2a}=3\Leftrightarrow\dfrac{c+2a}{ca}=\dfrac{1}{3}\Leftrightarrow\dfrac{1}{a}+\dfrac{2}{c}=\dfrac{1}{3}\)
Cộng vế theo vế \(\Leftrightarrow\dfrac{3}{a}+\dfrac{3}{b}+\dfrac{3}{c}=\dfrac{2}{3}+\dfrac{3}{4}+\dfrac{1}{3}=\dfrac{7}{4}\)
\(\Leftrightarrow3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{7}{4}\\ \Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{7}{12}\\ \Leftrightarrow\dfrac{ab+bc+ca}{abc}=\dfrac{7}{12}\\ \Leftrightarrow T=\dfrac{12}{7}\)
Bài 2:
a) Ta có: \(A=\dfrac{4}{n-1}+\dfrac{6}{n-1}-\dfrac{3}{n-1}\)
\(=\dfrac{4+6-3}{n-1}\)
\(=\dfrac{7}{n-1}\)
Để A là số tự nhiên thì \(7⋮n-1\)
\(\Leftrightarrow n-1\inƯ\left(7\right)\)
\(\Leftrightarrow n-1\in\left\{1;7\right\}\)
hay \(n\in\left\{2;8\right\}\)
Vậy: \(n\in\left\{2;8\right\}\)
ta có B=2n+9/n+2-3n+5n+1/n+2=4n+10/n+2 Để B là STN thì 4n+10⋮n+2 4n+8+2⋮n+2 4n+8⋮n+2 ⇒2⋮n+2 n+2∈Ư(2) Ư(2)={1;2} Vậy n=0
\(a)\,\,A=\dfrac{13}{21} \Leftrightarrow \dfrac{2n+3}{4n+1}=\dfrac{13}{21} \\ \Leftrightarrow 21(2n+3)=13(4n+1)\\\Leftrightarrow 42n+63=52n+13\\\Leftrightarrow 42n-52n=13-63 \\\Leftrightarrow -10n=-50\\\Leftrightarrow n=(-50):(-10)\\\Leftrightarrow n=5\)
b: =>\(\dfrac{2}{2}+\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{n\left(n+1\right)}=\dfrac{200}{101}\)
=>\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{n\left(n+1\right)}=\dfrac{100}{101}\)
=>1-1/2+1/2-1/3+...+1/n-1/n+1=100/101
=>1-1/(n+1)=100/101
=>1/(n+1)=1/101
=>n+1=101
=>n=100
