
2个回答
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原式
=(1-x/3-x)²/(3-2x-x²/9-x²)³*(x²/x²-2x-3)
=(x-1)²/(x-3)² × (x+3)³(x-3)³/(x-1)³(x+3)³ × x²/(x+1)(x-3)
= 1/(x-1)× x²/(x+1)
=x²/(x+1)(x-1)
=x²/(x²-1)
=(1-x/3-x)²/(3-2x-x²/9-x²)³*(x²/x²-2x-3)
=(x-1)²/(x-3)² × (x+3)³(x-3)³/(x-1)³(x+3)³ × x²/(x+1)(x-3)
= 1/(x-1)× x²/(x+1)
=x²/(x+1)(x-1)
=x²/(x²-1)
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