这个题...可真难已知:xˇ2-5x-1994=0求代数式[(x-2)ˇ4+(x-1)ˇ2-1]/[(x-1)(x-2)]的值
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这个题...可真难已知:xˇ2-5x-1994=0求代数式[(x-2)ˇ4+(x-1)ˇ2-1]/[(x-1)(x-2)]的值
这个题...可真难
已知:xˇ2-5x-1994=0
求代数式[(x-2)ˇ4+(x-1)ˇ2-1]/[(x-1)(x-2)]的值
这个题...可真难已知:xˇ2-5x-1994=0求代数式[(x-2)ˇ4+(x-1)ˇ2-1]/[(x-1)(x-2)]的值
(x-2)^4+(x-1)^2-1=(x-1)^2+[(x-2)^2+1][(x-2)^2-1]
=(x-1)^2+[(x-2)^2+1][(x-2)+1][(x-2)-1]
=(x-1)^2+[(x-2)^2+1](x-1)(x-3)
=(x-1)[x-1+x(x-2)^2+x-3(x-2)^2-3]
=(x-1)[2x-4+x(x-2)^2-3(x-2)^2]
=(x-1)(x-2)[2+x(x-2)-3(x-2)]
=(x-1)(x-2)[2+x^2-2x-3x+6]
=(x-1)(x-2)[x^2-5x+8]
[(x-2)ˇ4+(x-1)ˇ2-1]/[(x-1)(x-2)]
=x^2-5x+8=1994+8=2002
xˇ2-5x-1994=0
x^2-5x=1994
[(x-2)ˇ4+(x-1)ˇ2-1]/[(x-1)(x-2)]
={(x-2)ˇ4+[(x-1)ˇ2-1]}/[(x-1)(x-2)]
=[(x-2)ˇ4+x(x-2)]/[(x-1)(x-2)]
=(x-2)[(x-2)^3+x]/[(x-1)(x-2)]
=[(x-2)^3+x]/(x-1)
=(x^3-6x^2+13x-8)/(x-1)
=(x-1)(x^2-5x+8)/(x-1)
=x^2-5x+8
=1994+8
=2002
[(x-2)ˇ4+(x-1)ˇ2-1]/[(x-1)(x-2)]
=[(x-2)ˇ4+(x-2)x)]/[(x-1)(x-2)]
=[(x-2)ˇ3+x]/(x-1)
=xˇ2-5x+8
=1994+8=2002
〔(x-2)^4+(x-1)^2-1〕=x^4-8x^3+25x^2-34x+16
(x-1)(x-2)=x^2-3x+2 两式相除=x^2-5x+8
因为x^2-5x-1994=0
x^2-5x=1994
所以〔(x-2)4+(x-1)2-1〕/(x-1)(x-2)的值=1994+8
=2002