复数z1=3+4i,z2=0,z3=c+(2c-6)i在复平面内对应的点分别为A,B,C.若角BAC是钝角,则实数c的取值范围是_____.答案c>=49/11且c不等于9已知实数m,n满足于m/(1+i)=1-ni,则双曲线mx^2-ny^2=1的离心率为______.答案是根
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![复数z1=3+4i,z2=0,z3=c+(2c-6)i在复平面内对应的点分别为A,B,C.若角BAC是钝角,则实数c的取值范围是_____.答案c>=49/11且c不等于9已知实数m,n满足于m/(1+i)=1-ni,则双曲线mx^2-ny^2=1的离心率为______.答案是根](/uploads/image/z/12227278-22-8.jpg?t=%E5%A4%8D%E6%95%B0z1%3D3%2B4i%2Cz2%3D0%2Cz3%3Dc%2B%282c-6%29i%E5%9C%A8%E5%A4%8D%E5%B9%B3%E9%9D%A2%E5%86%85%E5%AF%B9%E5%BA%94%E7%9A%84%E7%82%B9%E5%88%86%E5%88%AB%E4%B8%BAA%2CB%2CC.%E8%8B%A5%E8%A7%92BAC%E6%98%AF%E9%92%9D%E8%A7%92%2C%E5%88%99%E5%AE%9E%E6%95%B0c%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4%E6%98%AF_____.%E7%AD%94%E6%A1%88c%3E%3D49%2F11%E4%B8%94c%E4%B8%8D%E7%AD%89%E4%BA%8E9%E5%B7%B2%E7%9F%A5%E5%AE%9E%E6%95%B0m%2Cn%E6%BB%A1%E8%B6%B3%E4%BA%8Em%2F%281%2Bi%29%3D1-ni%2C%E5%88%99%E5%8F%8C%E6%9B%B2%E7%BA%BFmx%5E2-ny%5E2%3D1%E7%9A%84%E7%A6%BB%E5%BF%83%E7%8E%87%E4%B8%BA______.%E7%AD%94%E6%A1%88%E6%98%AF%E6%A0%B9)
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复数z1=3+4i,z2=0,z3=c+(2c-6)i在复平面内对应的点分别为A,B,C.若角BAC是钝角,则实数c的取值范围是_____.答案c>=49/11且c不等于9已知实数m,n满足于m/(1+i)=1-ni,则双曲线mx^2-ny^2=1的离心率为______.答案是根
复数z1=3+4i,z2=0,z3=c+(2c-6)i在复平面内对应的点分别为A,B,C.若角BAC是钝角,则实数c的取值范围是_____.
答案c>=49/11且c不等于9
已知实数m,n满足于m/(1+i)=1-ni,则双曲线mx^2-ny^2=1的离心率为______.
答案是根号三
复数z1=3+4i,z2=0,z3=c+(2c-6)i在复平面内对应的点分别为A,B,C.若角BAC是钝角,则实数c的取值范围是_____.答案c>=49/11且c不等于9已知实数m,n满足于m/(1+i)=1-ni,则双曲线mx^2-ny^2=1的离心率为______.答案是根
由题意得点A的坐标为(3.4),B点的坐标为(0.0),C点的坐标为(c.2c-6)
在三角形BAC中,由余弦定理得:cos∠BAC=(AB²+AC²-BC²)/2AB*AC
所以cos∠BAC=(98-22c)/10*(5c-4)(c-9)½
= ﹤0
所以98-22c﹤0即c>49/11,又因为分母不能为0,所以c不等于9.
(2)因为m/(1+i)=m(1-i)2=1-ni
所以m=2,n=1
将m=2,n=1代人双曲线中得:x²-y²=1,
所以离心率e=√3