已知x^2-5x-2010=0则分式[(x-2)^4+(x-1)^2-1]/(x-1)(x-2)的值为

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/05 15:04:04
已知x^2-5x-2010=0则分式[(x-2)^4+(x-1)^2-1]/(x-1)(x-2)的值为
x){}K+tM+t l v|tOPH3DHjꃙ`Z6ycMR>U/^h sjl !N46:dPPSdV"VeOi%jH ckmCm(d4ի @,:|4E cA5aPA m (ؐ6yvm

已知x^2-5x-2010=0则分式[(x-2)^4+(x-1)^2-1]/(x-1)(x-2)的值为
已知x^2-5x-2010=0则分式[(x-2)^4+(x-1)^2-1]/(x-1)(x-2)的值为

已知x^2-5x-2010=0则分式[(x-2)^4+(x-1)^2-1]/(x-1)(x-2)的值为
令a=x-2
x-1=a+1
原式=[a^4+(a+1)²-1]/a(a+1)
=(a^4+a²+2a)/a(a+1)
=a(a³+1+a+1)/a(a+1)
=[(a+1)(a²-a+1)+(a+1)]/(a+1)
=(a+1)(a²-a+1+1)/(a+1)
=a²-a+2
=(x-2)²-(x-2)+2
=x²-5x+8
=2010+8
=2018