Giải thích các bước giải:
a.Để $A=\dfrac{1-6n}{2n-3}\in Z$
$\to 1-6n\quad\vdots\quad 2n-3$
$\to 1-6n+9-9\quad\vdots\quad 2n-3$
$\to 1-(6n-9)-9\quad\vdots\quad 2n-3$
$\to 1-3(2n-3)-9\quad\vdots\quad 2n-3$
$\to 1-9\quad\vdots\quad 2n-3$
$\to -8\quad\vdots\quad 2n-3$
$\to 8\quad\vdots\quad 2n-3$
$\to 2n-3\in U(8)$
Mà $2n-3$ lẻ
$\to 2n-3\in\{1,-1\}$
$\to 2n\in\{4,2\}$
$\to n\in\{2,1\}$
b.Ta có:
$\dfrac{ab}{a+2b}=\dfrac32$
$\to \dfrac{a+2b}{ab}=\dfrac23$
$\to \dfrac1b+\dfrac2a=\dfrac23$
$\to\dfrac1b=\dfrac23-\dfrac2a$
Ta có:
$\dfrac{ca}{c+2a}=3$
$\to\dfrac{c+2a}{ca}=\dfrac13$
$\to\dfrac1a+\dfrac2c=\dfrac13$
$\to \dfrac2c=\dfrac13-\dfrac1a$
$\to \dfrac1c=\dfrac16-\dfrac1{2a}$
Ta có:
$\dfrac{bc}{b+2c}=\dfrac43$
$\to\dfrac{b+2c}{bc}=\dfrac43$
$\to\dfrac1c+\dfrac2b=\dfrac43$
$\to(\dfrac16-\dfrac1{2a})+2\cdot (\dfrac23-\dfrac2a)=\dfrac43$
$\to \dfrac{3}{2}-\dfrac{9}{2a}=\dfrac43$
$\to 9a-27=8a$
$\to a=27$
$\to \dfrac1b=\dfrac23-\dfrac2{27}\to b=\dfrac{27}{16}$
$\dfrac1c=\dfrac16-\dfrac1{2\cdot 27}\to c=\dfrac{27}{4}$
