x²+1＜ax+a分之x a不等于 解不等式

a≠0x²+1＜ax+x/a x²+1＜x(a+1/a) x=0时,0+1＜0恒不成立,所以x≠0如果x＞0,两边同除以x：x+1/x＜a+1/a∵x+1/x＞0,∴a+1/a＞0,∴a＞0x-a+1/x-1/a＜0(x-a)-(x-a)/(ax)＜0(x-a)(ax-1)/(ax)＜0(x-a)(x-1/a)＜0当0＜a＜1时,a＜x＜1/a当a=1时无解当a＞1时,1/a＜x＜a如果x＜0,两边同除以x：x+1/x＞a+1/a∵x+1/x＜0,∴a+1/a＜0,∴a＜0x-a+1/x-1/a＞0(x-a)-(x-a)/(ax)＞0(x-a)(ax-1)/(ax)＞0(x-a)(ax-1)＞0(x-a)(x-1/a)＜0当a＜-1时,a＜x＜1/a当a=-1时无解当-1＜a＜0时,1/a＜x＜a