(x分之x-1-x+1分之x-2)除以x^2+2X+1分之2x^2-X ,其中满足x^2-X-1=0 需要过程
1个回答
2013-10-17
展开全部
[(x-1)/x - (x-2)/(x+1)] ÷ (2x�0�5-x)/(x�0�5+2x+1)
=[(x+1)(x-1)-x(x-2)]/x(x+1) × (x�0�5+2x+1)/x(2x-1)
=(x�0�5-1-x�0�5+2x)/x(x+1) × (x+1)�0�5/x(2x-1)
=(2x-1)/x × (x+1)/x(2x-1)
= (x+1)/x�0�5
因为x�0�5-x-1=0
所以原式=(x+1)/x�0�5=x�0�5/x�0�5=1
=[(x+1)(x-1)-x(x-2)]/x(x+1) × (x�0�5+2x+1)/x(2x-1)
=(x�0�5-1-x�0�5+2x)/x(x+1) × (x+1)�0�5/x(2x-1)
=(2x-1)/x × (x+1)/x(2x-1)
= (x+1)/x�0�5
因为x�0�5-x-1=0
所以原式=(x+1)/x�0�5=x�0�5/x�0�5=1
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