设x^2=6x-7,则x^4-4x^3-3x^2+2x+29/x^2-6x+10=?
热心网友
设x^2=6x-7,则(x^4-4x^3-3x^2+2x+29)/(x^2-6x+10)=? x^2=6x-7,所以x^2-6x+7=0,因为(x^4-4x^3-3x^2+2x+29)/(x^2-6x+7)商(x^2+2x+2)余数为15,所以(x^4-4x^3-3x^2+2x+29)/(x^2-6x+10)=［(x^2-6x+7)(x^2+2x+2)+15］/［(x^2-6x+7)+3］=15/3=5
热心网友
x^4-4x^3-3x^2+2x+29=(x^2)^2-x^2(4x+3)+2x+29=(6x-7)^2-(6x-7)(4x+3)+2x+29=12x^2-72x+105=12(6x-7)-72x+105=21x^2-6x+10=6x-7-6x+10=3,所以(x^4-4x^3-3x^2+2x+29)/(x^2-6x+10)=21/3=7