练习册

  • 练习册
  • 试题
    21．解:(1)令.得 解得 令.得 ∴ A B C ·················· (2)∵OA=OB=OC= ∴BAC=ACO=BCO= ∵AP∥CB. ∴PAB= 过点P作PE轴于E.则APE为等腰直角三角形 令OE=.则PE= ∴P ∵点P在抛物线上 ∴ 解得. ∴PE=······································································································ 4分) ∴四边形ACBP的面积=AB•OC+AB•PE =······················································································· 6分) (3)． 假设存在 ∵PAB=BAC = ∴PAAC ∵MG轴于点G. ∴MGA=PAC = 在Rt△AOC中.OA=OC= ∴AC= 在Rt△PAE中.AE=PE= ∴AP= ······························································ 7分) 设M点的横坐标为.则M ①点M在轴左侧时.则 (ⅰ) 当AMG PCA时.有= ∵AG=.MG= 即 解得 (ⅱ) 当MAG PCA时有= 即 解得: ∴M ···································································································· ② 点M在轴右侧时.则 (ⅰ) 当AMG PCA时有= ∵AG=.MG= ∴ 解得 ∴M (ⅱ) 当MAGPCA时有= 即 解得: ∴M ∴存在点M.使以A.M.G三点为顶点的三角形与PCA相似 M点的坐标为.. ························································· 【查看更多】

     

    题目列表(包括答案和解析)

    方程|x－1|＋|x－2|＝3的解法

    　　要解方程|x－1|＋|x－2|＝3，就应设法化去绝对值符号，而要去掉绝对值符号就要判断x－1，x－2是正数还是负数？

    　　解：令x－1＝0得x＝1，令x－2＝0得x＝2，分x＜1、1≤x＜2、x≥2三种情况化去绝对值符号．

    　　当x＜1时，原方程可化为(1－x)＋(2－x)＝3，

    　　解得x＝0．

    　　当1≤x＜2时，原方程可化为(x－1)＋(2－x)＝3，从而1＝3，这不可能．说明x的值不可能是1≤x＜2．

    　　当x≥2时，原方程可化为x－1＋x－2＝3，解得x＝3．

    　　∴原方程的解为x＝0和x＝3．

    读了上面的内容你有何启发？你能求方程|x＋1|＋|x＋3|＝5的解吗？试试看．

    查看答案和解析>>


    同步练习册答案