实数a与b满足等式a^2b^2/a^4-2b^4=1,则a^2-b^2/19a^2+96b^2的值是多少?
问题描述:
实数a与b满足等式a^2b^2/a^4-2b^4=1,则a^2-b^2/19a^2+96b^2的值是多少?
答
a^2b^2/a^4-2b^4=1a^2b^2=a^4-2b^4a^4-2b^4-a^2b^2=0(a^2-2b^2)(a^2+b^2)=0a^2-2b^2=0或a^2+b^2=0所以a^2=2b^2或-b^2当a^2=2b^2时a^2-b^2/19a^2+96b^2=(2b^2-b^2)/(38b^2+96b^2)=1/124当a^2=-b^2时a^2-b^2/19a^2+96b...