Giải thích các bước giải:
Bài 1:
a)đkxđk : $a-1\ne 0$
b)$M=(1+\dfrac{a}{a^2+1}):(\dfrac{1}{a-1}-\dfrac{2a}{a^3-a^2+a-1})$
$\to M=\dfrac{a^2+1+a}{a^2+1}:(\dfrac{1}{a-1}-\dfrac{2a}{(a-1)(a^2+1)})$
$\to M=\dfrac{a^2+1+a}{a^2+1}:\dfrac{a^2+1-2a}{(a-1)(a^2+1)}$
$\to M=\dfrac{a^2+1+a}{a^2+1}:\dfrac{(a-1)^2}{(a-1)(a^2+1)}$
$\to M=\dfrac{a^2+a+1}{a^2+1}:\dfrac{a-1}{a^2+1}$
$\to M=\dfrac{a^2+a+1}{a^2+1}.\dfrac{a^2+1}{a-1}$
$\to M=\dfrac{a^2+a+1}{a-1}$
c.Để $M\in Z\to a^2+a+1\quad\vdots\quad a-1$
$\to a^2-1+a-1+3\quad\vdots\quad a-1$
$\to (a-1)(a+1)+(a-1)+3\quad\vdots\quad a-1$
$\to 3\quad\vdots\quad a-1$
$\to a-1\in\{1,3,-1,-3\}\to a\in\{2,4,0,-2\}$
Bài 2 :
Để $f(x)$ chia $g(x)$ dư 4
$\to f(3)+4=0\to 2.3^2+3a+1+4=0\to a=\dfrac{-23}3$
