x,y,z是三个有理数,若x<y,x+y=0,且xyz>0. (1)试判断x,y,z的正负性; x,y,z是三个有理数,若x<y,x+y=o,且xyz>o,...
x\uff0cy\uff0cz\u662f\u4e09\u4e2a\u6709\u7406\u6570\uff0c\u82e5x<y\uff0cx+y=0\uff0c\u4e14xyz>0. 1\u3001\u8bd5\u5224\u65adx\uff0cy\uff0cz\u7684\u6b63\u8d1f\u6027; 21\uff1ax\uff1c0 y\uff1e0 z\uff1e0 \u56e0\u4e3ax\uff0by\uff1d0\u4e14x\uff1cy\u6240\u4ee5x\uff1c0 y\uff1e0\u4e14X\u7684\u7edd\u5bf9\u503c\u4e00\u5b9a\u7b49\u4e8ey xy\uff1c0 \u53c8xyz\uff1e0\u6240\u4ee5Z\uff1c0
2\uff1a\u56e0\u4e3ax\uff1c0 y\uff1e0 \u6240\u4ee5x\uff0dy\uff1c0 Z\uff1c0 \u6240\u4ee5x\uff0bZ\uff1c0\u6240\u4ee5\uff08x\uff0bz\uff09\uff08x\uff0dy\uff09\uff1e0
x+y=0\u89e3\u5f97x=-y
\u90a3\u4e48\uff1ax=-y<y,\u8bf4\u660ey\u662f\u6b63\u53f7\u7684\uff0cx\u662f\u8d1f\u53f7\u7684\u3002
xyz>0\u53ea\u80fd\u77e5\u9053z\u4e3a\u8d1f\u53f7\uff0c\u56e0\u4e3axy\u662f\u8d1f\u53f7\u3002
x\uff0cz\u90fd\u662f\u8d1f\u6570\uff0c\u76f8\u52a0\u4e5f\u662f\u8d1f\u6570
x=-y\u89e3\u5f97x-y=-y-y=-2y
y\u662f\u6b63\u6570\uff0c-2y\u5c31\u662f\u8d1f\u6570
\u4e24\u4e2a\u8d1f\u6570\u76f8\u4e58\u5f97\u5230\u6b63\u6570
2)符号为正号
过程:∵x<
0 ,y>0 ,z<0 .∴x+z<0,x-y<0.∴(x+z)(x-y)>0符号为正
懂了么?
如果还有问题的话欢迎追问
望采纳谢谢~
只给你解题思路:
(1)x+y=0,x与y异号或为零,因x<y,则x,y各自正负就可判断出来。又因xyz>0,则z的正负也可以判断出来了。
(2)依上问,依次判断x,z,x,-y的正负,就可判断两个因式(x+z)和(x-y)的正负,答案自然出来了。
X是负数,y是正数z是正数
第二题不会,求采纳
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