解一元二次不等式：①x²+2x-3≤0

题目

解一元二次不等式：①x²+2x-3≤0
①x²+2x-3≤0
②x-x²+6＜0
③4x²+4x+1≥0
④x²-6x+9≤0
⑤-4+x-x²＜0

求过程解答,谢谢O(∩_∩)O了!

答案

①x²+2x-3≤0(x+3)(x-1)≤0-3≤x≤1②x-x²+6＜0-x²+x+6＜0x²-x-6＞0(x+2)(x-3)＞0x＜-2或x＞3③4x²+4x+1≥0(2x+1)²≥0x≠-1/2④x²-6x+9≤0（x-3)²≤0x=3⑤-4+x-x²＜0x...