Cho biểu thức A=($\frac{x^{2}-16}{x-4}$-1):($\frac{x-2}{x-3}$+$\frac{x+3}{x+1}$+$\frac{x+2- x^{2}}{ x^{2}-2x-3}$ ) a) Rút gọn A b) Tìm x nguyên để $\f

Giải thích các bước giải:

a.Ta có:

$A=(\dfrac{x^2-16}{x-4}-1):(\dfrac{x-2}{x-3}+\dfrac{x+3}{x+1}+\dfrac{x+2-x^2}{x^2-2x-3})$

$\to A=(\dfrac{(x-4)(x+4)}{x-4}-1):(\dfrac{x-2}{x-3}+\dfrac{x+3}{x+1}-\dfrac{x^2-x-2}{x^2-3x+x-3})$

$\to A=(x+4-1):(\dfrac{x-2}{x-3}+\dfrac{x+3}{x+1}-\dfrac{x^2-x-2}{(x-3)(x+1)})$

$\to A=(x+3):(\dfrac{x-2}{x-3}+\dfrac{x+3}{x+1}-\dfrac{x^2-2x+x-2}{(x-3)(x+1)})$

$\to A=(x+3):(\dfrac{x-2}{x-3}+\dfrac{x+3}{x+1}-\dfrac{(x-2)(x+1)}{(x-3)(x+1)})$

$\to A=(x+3):(\dfrac{x-2}{x-3}+\dfrac{x+3}{x+1}-\dfrac{x-2}{x-3})$

$\to A=(x+3):\dfrac{x+3}{x+1}$

$\to A=(x+3)\cdot \dfrac{x+1}{x+3}$

$\to A=x+1$

b.Ta có:

$B=\dfrac{A}{x^2+x+1}$

$\to B=\dfrac{x+1}{x^2+x+1}$

$\to B(x^2+x+1)=x+1$

$\to Bx^2+Bx+B=x+1$

$\to Bx^2+x(B-1)+B-1=0(*)$

Nếu $B=0\to x+1=0\to x=-1$

Nêu $B\ne 0\to (*)$ là phương trình bậc $2$ ẩn $x$

Để phương trình có nghiệm

$\to \Delta\ge 0$

$\to (B-1)^2-4B(B-1)\ge 0$

$\to (B-1)(B-1-4B)\ge 0$

$\to (B-1)(-3B-1)\ge 0$

$\to (B-1)(3B+1)\le 0$

$\to -\dfrac13\le B\le 1$

Mà $B\in Z, B\ne 0$

$\to B=1$

$\to \dfrac{x+1}{x^2+x+1}=1\to x+1=x^2+x+1\to x^2=0\to x=0$

$\to x\in\{0,-1\}$