Question

第17题怎么做？求解答，谢谢

第17题

Answer 1

x=2-√2
[(x^2-4)/x] .[ (x+2)/(x^2-2x) - (x-1)/(x^2-4x+4) ]/ [ (x^2-2x-8)/x^2]
=[(x-2)(x+2)/x] .[ (x+2)/(x(x-2)) - (x-1)/(x-2)^2 ]/ [ (x-4)(x+2)/x^2]
=[ (x+2)/x - (x-1)/(x-2)/ [ (x-4)/x ]

=[ (x+2) - x(x-1)/(x-2)]/ (x-4)
=[ (x+2)(x-2) - x(x-1) ]/[ (x-4)(x-2)]
=( x-4)/[ (x-4)(x-2)]
=1/(x-2)
=1/(-√2)
=-√2/2

Answer 2

[(x^2-4)/x]*[(x+2)/(x^2-2x)-(x-1)/(x^2-4x+4)]÷[(x^2-2x-8)/x^2]
=[(x+2)(x-2)/x]*[(x+2)/x(x-2)-(x-1)/(x-2)^2]*[(x^2)/(x-4)(x+2)]
=[x(x-2)/(x-4)]*[(x^2-4-x^2+x)/x(x-2)^2]
=(x-4)/(x-4)(x-2)
=1/(x-2)
将x=2-√2代入

原式=1/(2-√2-2)=-√2/2

Answer 3

[(x^2-4)/x]*[(x+2)/(x^2-2x)-(x-1)/(x^2-4x+4)]÷[(x^2-2x-8)/x^2]
=[(x+2)(x-2)/x]*[(x+2)/x(x-2)-(x-1)/(x-2)^2]*[(x^2)/(x-4)(x+2)]
=[x(x-2)/(x-4)]*[(x^2-4-x^2+x)/x(x-2)^2]
=(x-4)/(x-4)(x-2)
=1/(x-2)
将x=2-√2代入

原式=1/(2-√2-2)=-√2/2