已知x+1/x=2,求分式(x^2+2x+1)/(4x^2-7x+4)的值
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已知x+1/x=2,求分式(x^2+2x+1)/(4x^2-7x+4)的值
已知x+1/x=2,求分式(x^2+2x+1)/(4x^2-7x+4)的值
已知x+1/x=2,求分式(x^2+2x+1)/(4x^2-7x+4)的值
(x^2+2x+1)/(4x^2-7x+4)
分子分母同时除以x得
原式=(x+2+1/x)/(4x-7+4/x)
=[(x+1/x)+2]/[4(x+1/x)-7]
=4/(8-7)
=4
x+1/x=2
=>x^2-2x+1=0
=>(x-1)^2=0
=>x=1
(x^2+2x+1)/(4x^2-7x+4)
=(x+1)^2/[4(x-1)^2+x]
=4/1
=4