Giải thích các bước giải:
a.Ta có:
$M=(1+\dfrac{a}{a^2+1}):(\dfrac{1}{a-1}-\dfrac{2a}{a^3+a-a^2-1})$
$\to M=\dfrac{a^2+1+a}{a^2+1}:(\dfrac{1}{a-1}-\dfrac{2a}{a(a^2+1)-(a^2+1)})$
$\to M=\dfrac{a^2+a+1}{a^2+1}:(\dfrac{a^2+1}{(a^2+1)(a-1)}-\dfrac{2a}{(a-1)(a^2+1)})$
$\to M=\dfrac{a^2+a+1}{a^2+1}:\dfrac{a^2+1-2a}{(a-1)(a^2+1)}$
$\to M=\dfrac{a^2+a+1}{a^2+1}:\dfrac{(a-1)^2}{(a-1)(a^2+1)}$
$\to M=\dfrac{a^2+a+1}{a^2+1}:\dfrac{a-1}{a^2+1}$
$\to M=\dfrac{a^2+a+1}{a^2+1}\cdot\dfrac{a^2+1}{a-1}$
$\to M=\dfrac{a^2+a+1}{a-1}$
b.Để $M=7$
$\to \dfrac{a^2+a+1}{a-1}=7$
$\to a^2+a+1=7(a-1)$
$\to a^2-6a+8=0$
$\to (a-4)(a-2)=0$
$\to a\in\{4,2\}$
c.Ta có:
$M(a-1)=a^2+a+1=(a+\dfrac12)^2+\dfrac34\ge\dfrac34$
Dấu = xảy ra khi $a=-\dfrac12$
d.Để $M<1$
$\to \dfrac{a^2+a+1}{a-1}<1$
$\to \dfrac{a^2+a+1}{a-1}-1<0$
$\to \dfrac{a^2+2}{a-1}<0$
Mà $a^2+2>0\to a-1<0\to a<1$
e.Để $M\in Z$
$\to \dfrac{a^2+a+1}{a-1}\in Z$
$\to \dfrac{(a^2-a)+(2a-2)+3}{a-1}\in Z$
$\to \dfrac{a(a-1)+2(a-1)+3}{a-1}\in Z$
$\to a+2+\dfrac3{a-1}\in Z$
Mà $a\in Z$
$\to a-1\in U(3)$
$\to a-1\in\{1,3,-1,-3\}$
$\to a\in\{2,4,0,-2\}$
