题目

解不等式≥7. 答案：思路解析：移项、通分后转化为一元二次不等式（组）或高次不等式求解.解法一：原不等式化为≥0,它等价于(x-)(x+)≤0的解为-≤x≤.又∵x≠1,∴原不等式的解集为{x|-≤x＜1或1＜x≤}.解法二：原不等式可化为≥0.它等价于由简易标根法易知其解集为{x|-≤x＜1或1＜x≤.误区警示    解分式不等式切忌不加讨The reason _____he was absent was _____ he was ill and unable to go to school.A．which; thatB．why; becauseC．which; becauseD．why; that