Question

a1+a1q的三次方=10 a1q+a1q的四次方=20 怎么解得

Answer 1

a1*q^4-a1=a1(q^4-1)=a1(q^2+1)(q^2-1)=15 a1*q^3-a1*q=a1*q(q^2-1)=6由已知可知q不等于正负1,所以q^2-1不等于0 两等式相除,左边除以左边,右边除以右边可得(q^2+1)/q=5/2 即 2q^2-5q+2=0 即 （2q-1)(q-2)=0...