Đáp án:
A= $\frac{2x^2(6-x^3)(2x+1)(2x-1)}{2x^(2-x)(x²-4)}$$\frac{x-2}{2x^2-x}$
Giải thích các bước giải:
A=( $\frac{x-2}{x+2}$ - $\frac{4²}{x²÷x^4}$ - $\frac{2+x}{x-2}$ ) ×$\frac{x-2}{2x^2-x}$
MTC : (x²-4)(x²-x^4)
A= ( $\frac{(x-2)(x-2)(x^2-x^4)}{(x²-4)(x²÷x^4)}$ - $\frac{4x²(x²-4)}{(x²-4)(x²÷x^4)}$ -$\frac{(2+x)(x-2)(x^2÷x^4)}{(x²-4)(x²÷x^4)}$) ×$\frac{x-2}{2x^2-x}$
A=( $\frac{x^4-x^6-2x^3+2x^5+4x^2-4x^4}{(x²-4)(x²÷x^4)}$ - $\frac{4x^4-16x^2}{(x²-4)(x²÷x^4)}$ - $\frac{(x²-4)(x²÷x^4)}{(x²-4)(x²÷x^4)}$) ×$\frac{x-2}{2x^2-x}$
A= ($\frac{x^4-x^6-2x^3+2x^5+4x^2-4x^4-4x^4+16x^2-x^4+x^6+4x^2-4x^4}{(x²-4)(x²÷x^4)}$) ×$\frac{x-2}{2x^2-x}$
A=( $\frac{2x^5+24x^2-2x^3-12x^4}{(x²-4)(x²÷x^4)}$) ×$\frac{x-2}{2x^2-x}$
A=( $\frac{(24x^2-12x^4)+(2x^5-2x^3)}{(x²-4)(x²÷x^4)}$) ×$\frac{x-2}{2x^2-x}$
A= ($\frac{12x^2(2x^2-1)+2x^3(2x^2-1)}{(x²-4)(x²÷x^4)}$) ×$\frac{x-2}{2x^2-x}$
A= ($\frac{(12x^2+2x^3)(2x^2-1)}{(x²-4)(x²÷x^4)}$) ×$\frac{x-2}{2x^2-x}$
A=( $\frac{2x^2(6-x^3)(2x+1)(2x-1)}{(x²-4)(x²÷x^4)}$) $\frac{2x^2(6-x^3)(2x+1)(2x-1)}{(x²-4)(x^-2)}$× $\frac{x-2}{2x^(2-x)}$
A= $\frac{2x^2(6-x^3)(2x+1)(2x-1)}{(x²-4)}$× $\frac{1}{2x^(2-x)}$
A= $\frac{2x^2(6-x^3)(2x+1)(2x-1)}{2x^(2-x)(x²-4)}$$\frac{x-2}{2x^2-x}$
Vậy A= $\frac{2x^2(6-x^3)(2x+1)(2x-1)}{2x^(2-x)(x²-4)}$$\frac{x-2}{2x^2-x}$
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