Câu 1 : cho biểu thức : P = ( 2+ x / 2- x – 4x^2 / x^2 + 4 – 2- x / 2+x ) : x^2 – 3x / 2x^2 – x^3 a) tìm điều kiện xác định và rút gọn P

Giải thích các bước giải:

a.Ta có hàm số xác định

$\to \begin{cases}2-x\ne 0\\ 2+x\ne 0\\x^2-4\ne 0\\ 2x^2-x^3\ne 0\\ x^2-3x\ne 0\end{cases}$

$\to \begin{cases}x\ne 2\\ x\ne -2\\ x\ne \pm2\\ x^2(2-x)\ne 0\\ x(x-3)\ne 0\end{cases}$

$\to \begin{cases}x\ne 2\\ x\ne -2\\ x\notin \{0,2\}\\ x\notin \{0,3\}\end{cases}$

$\to x\notin\{0,2,3,-2\}$

Ta có:

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

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

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

$\to P=\dfrac{(2+x)^2+4x^2-(2-x)^2}{(2+x)(2-x)}\cdot \dfrac{x(2-x)}{x-3}$

$\to P=\dfrac{4x^2+8x}{(2+x)(2-x)}\cdot \dfrac{x(2-x)}{x-3}$

$\to P=\dfrac{4x(x+2)}{(2+x)(2-x)}\cdot \dfrac{x(2-x)}{x-3}$

$\to P=\dfrac{4x^2}{x-3}$

b.Ta có:

$|x-5|=3\to x-5=3\to x=8$ hoặc $x-5=-3\to x=2$ loại vì $x\ne 2$

$\to P=\dfrac{4\cdot 8^2}{8-3}=\dfrac{256}{5}$

c.Ta có:

$Q=\dfrac1P$

$\to Q=\dfrac{x-3}{4x^2}$

$\to Q=\dfrac1{4x}-\dfrac3{4x^2}$

$\to Q=-(\dfrac3{4x^2}-\dfrac1{4x})$

$\to Q=-3(\dfrac1{4x^2}-\dfrac1{12x})$

$\to Q=-3((\dfrac1{2x})^2-2\cdot\dfrac1{2x}\cdot \dfrac{1}{12}+(\dfrac{1}{12})^2-(\dfrac{1}{12})^2)$

$\to Q=-3((\dfrac1{2x}-\dfrac{1}{12})^2-(\dfrac{1}{12})^2)$

$\to Q\le -3(0-(\dfrac{1}{12})^2)$

$\to Q\le \dfrac1{48}$

Dấu = xảy ra khi $\dfrac1{2x}-\dfrac1{12}=0\to x=6$