Question

①若2x²－3y＋z＝0,3x－2y－6z＝0,求x²＋y²＋z²除以2x

Answer 1

1）令y=0,2x²＋z＝0,3x－6z＝0,得x=-1/4,z=-1/8,代入原式=5/7

2）（x^10+x^6+x^4+1)/(x^10+x^8+x^2+1)=(x^6+1)(x^4+1)/(x^8+1)(x^2+1)=(x^4-x^2+1)(x^4+1)/(x^8+1)

上下都除以x^4

(x-1/x)^2=5 x²+1/x²=7 x^4+1/x^4=47

原式=42/47