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∵x²+X-1=0
∴x²+x=1 x²=1-x
X(1减1-X分之2)除以(X+1)减 X的平方-2X+1分之X(X的平方-1)
=x[1-2/(1-x)]÷(x+1)-x(x²-1)/(x²-2x+1)
=x[(1-x-2)/(1-x)]÷(x+1)-x(x-1)(x+1)/(x-1)²
=x[(-x-1)/(1-x)]÷(x+1)-x(x+1)/(x-1)
=x[(x+1)/(x-1)(x+1)]-(x²+x)/(x-1)
=x/(x-1)-1/(x-1)
=(x-1)/(x-1)
=1
∴x²+x=1 x²=1-x
X(1减1-X分之2)除以(X+1)减 X的平方-2X+1分之X(X的平方-1)
=x[1-2/(1-x)]÷(x+1)-x(x²-1)/(x²-2x+1)
=x[(1-x-2)/(1-x)]÷(x+1)-x(x-1)(x+1)/(x-1)²
=x[(-x-1)/(1-x)]÷(x+1)-x(x+1)/(x-1)
=x[(x+1)/(x-1)(x+1)]-(x²+x)/(x-1)
=x/(x-1)-1/(x-1)
=(x-1)/(x-1)
=1
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