已知:a+b=6,(ab)的负一次方=-1/7,求a-b/3ab ×(a²-b²/a²+2ab+
问题描述:
已知:a+b=6,(ab)的负一次方=-1/7,求a-b/3ab ×(a²-b²/a²+2ab+
²)的值
答
a-b/3ab ×(a²-b²/a²+2ab+²)=a-b/3ab*【(a+b)(a-b)/(a+b)(a+b)】=(a-b)(a-b)/3ab(ab)的负一次方=-1/7则:ab=-7(a-b)(a-b)=(a+b)(a+b)-4ab=36+28=64原式=64/3*(-7)=-64/...