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1、高二数学（理）基础训练（12）1、不等式的解集是( )32x(A) (B) (C) (D) )32,()32,(), 0( U)0 ,32(), 0( U)0 ,32(2、不等式|x-1|+|x+2|的解集为( )5(A). (B). , 22,U , 21,U(C). ( D). , 32,U , 23,U3、已知a,b,c是正实数,且a+b+c=1,则的最小值为( )cba111A.3 B. 6 C. 9 D. 12 4、若复数z满足3(1)1zz i,则2zz的值等于( )A1 B0 C1 D13 22i5、给出下列命题 (1)实数的共轭复数一定是实数;(2)满足2zizi 的复数z的轨
2、迹是椭圆;(3)若2,1mZ i ,则1230;mmmmiiii其中正确命题的序号是( )A.(1) B.(2)(3) C.(1)(3) D.(1)(4)6、若 12zai, 234zi，且12z z为纯虚数，则实数a的值为 7、计算221111iiii_。8、函数 y=的最大值为 ；xx52159、建造一个容积为 18 m3，深为 2 m 的长方体无盖水池，如果池底和池壁每平方米的造价 分别为 200 元和 150 元，那么池的最低造价为：_.10、设，若，求的最大值为：_。, a bR225ab2ab11、解不等式xxx3423212、已知复数z满足2z ，2z的虚部是 2（1）求复数z；
3、（2）设22zzzz，在复平面上的对应点分别为ABC，求ABC的面积基础训练(12)参考答案1-5 BDCCC6、8 3 7、18、9、5400 元 10、529211、256| xx12、解：（1）设()zabi abR，则2222zababi，由题意得222ab且22ab ， 解得1ab或1ab ， 因此1zi 或1zi （2）当1zi 时，22zi，21zzi ，所以得(11)(0 2)(11)ABC，所以1ABCS当1zi 时，22zi，213zzi ，所以得( 11)(0 2)( 13)ABC，所以1ABCS9JWKffwvG#tYM*Jg&6a*CZ7H$dq8KqqfHVZFed
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