麻烦大家看下这道题:复数z=x+yi(x, y∈R)满足|z-4i|=|z+2|,则2^x+4^y的最小值是由|z-4i|=|z+2|,得:|x+yi-4i|=|x+yi+2|,|x+(y-4)i|=|(x+2)+yi|,即√[x^2+(y-4)^2]=√[(x+2)^2+y^2],x^2+(y-4)^2=(x+2)^2+y^2,化简,得:x+2y=3.又2^x>0,
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/17 08:02:33
![麻烦大家看下这道题:复数z=x+yi(x, y∈R)满足|z-4i|=|z+2|,则2^x+4^y的最小值是由|z-4i|=|z+2|,得:|x+yi-4i|=|x+yi+2|,|x+(y-4)i|=|(x+2)+yi|,即√[x^2+(y-4)^2]=√[(x+2)^2+y^2],x^2+(y-4)^2=(x+2)^2+y^2,化简,得:x+2y=3.又2^x>0,](/uploads/image/z/10409426-26-6.jpg?t=%E9%BA%BB%E7%83%A6%E5%A4%A7%E5%AE%B6%E7%9C%8B%E4%B8%8B%E8%BF%99%E9%81%93%E9%A2%98%EF%BC%9A%E5%A4%8D%E6%95%B0z%3Dx%2Byi%28x%2C+y%E2%88%88R%29%E6%BB%A1%E8%B6%B3%7Cz-4i%7C%3D%7Cz%2B2%7C%2C%E5%88%992%5Ex%2B4%5Ey%E7%9A%84%E6%9C%80%E5%B0%8F%E5%80%BC%E6%98%AF%E7%94%B1%7Cz-4i%7C%3D%7Cz%2B2%7C%2C%E5%BE%97%EF%BC%9A%7Cx%2Byi-4i%7C%3D%7Cx%2Byi%2B2%7C%2C%7Cx%2B%28y-4%29i%7C%3D%7C%28x%2B2%29%2Byi%7C%2C%E5%8D%B3%E2%88%9A%5Bx%5E2%2B%28y-4%29%5E2%5D%3D%E2%88%9A%5B%28x%2B2%29%5E2%2By%5E2%5D%2Cx%5E2%2B%28y-4%29%5E2%3D%28x%2B2%29%5E2%2By%5E2%2C%E5%8C%96%E7%AE%80%2C%E5%BE%97%EF%BC%9Ax%2B2y%3D3.%E5%8F%882%5Ex%3E0%2C)
麻烦大家看下这道题:复数z=x+yi(x, y∈R)满足|z-4i|=|z+2|,则2^x+4^y的最小值是由|z-4i|=|z+2|,得:|x+yi-4i|=|x+yi+2|,|x+(y-4)i|=|(x+2)+yi|,即√[x^2+(y-4)^2]=√[(x+2)^2+y^2],x^2+(y-4)^2=(x+2)^2+y^2,化简,得:x+2y=3.又2^x>0,
麻烦大家看下这道题:复数z=x+yi(x, y∈R)满足|z-4i|=|z+2|,则2^x+4^y的最小值是
由|z-4i|=|z+2|,得:|x+yi-4i|=|x+yi+2|,|x+(y-4)i|=|(x+2)+yi|,即√[x^2+(y-4)^2]=√[(x+2)^2+y^2],x^2+(y-4)^2=(x+2)^2+y^2,化简,得:x+2y=3.又2^x>0, 2^2y>0, 所以2^x+4^y=2^x+2^2y>=2*√(2^x*2^2y)=2*√2^3=4√2. (为什么2^x+2^2y>=2*√(2^x*2^2y),它为什么要大于2倍的它呢?)当且仅当2^x=2^2y,即x=2y,x=3/2,y=3/4时,取等号.故所求的2^x+4^y的最小值是:4√2.
麻烦大家看下这道题:复数z=x+yi(x, y∈R)满足|z-4i|=|z+2|,则2^x+4^y的最小值是由|z-4i|=|z+2|,得:|x+yi-4i|=|x+yi+2|,|x+(y-4)i|=|(x+2)+yi|,即√[x^2+(y-4)^2]=√[(x+2)^2+y^2],x^2+(y-4)^2=(x+2)^2+y^2,化简,得:x+2y=3.又2^x>0,