设实数x,y,m,n满足x^2+y^2=3,m^2+n^2=1,若a≥mx+ny恒成立,求a的取值范围(x-m)^2≥0(x^2+m^2)/2≥xm(y-n)^2≥0(y^2+n^2)/2≥yn(x^2+m^2+y^2+n^2)/2≥xm+yn2≥xm+yna≥xm+yna≥2这个做法为什么是错误的?正确解法是什么?
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![设实数x,y,m,n满足x^2+y^2=3,m^2+n^2=1,若a≥mx+ny恒成立,求a的取值范围(x-m)^2≥0(x^2+m^2)/2≥xm(y-n)^2≥0(y^2+n^2)/2≥yn(x^2+m^2+y^2+n^2)/2≥xm+yn2≥xm+yna≥xm+yna≥2这个做法为什么是错误的?正确解法是什么?](/uploads/image/z/14996883-3-3.jpg?t=%E8%AE%BE%E5%AE%9E%E6%95%B0x%2Cy%2Cm%2Cn%E6%BB%A1%E8%B6%B3x%5E2%2By%5E2%3D3%2Cm%5E2%2Bn%5E2%3D1%2C%E8%8B%A5a%E2%89%A5mx%2Bny%E6%81%92%E6%88%90%E7%AB%8B%2C%E6%B1%82a%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4%28x-m%29%5E2%E2%89%A50%28x%5E2%2Bm%5E2%29%2F2%E2%89%A5xm%28y-n%29%5E2%E2%89%A50%28y%5E2%2Bn%5E2%29%2F2%E2%89%A5yn%28x%5E2%2Bm%5E2%2By%5E2%2Bn%5E2%29%2F2%E2%89%A5xm%2Byn2%E2%89%A5xm%2Byna%E2%89%A5xm%2Byna%E2%89%A52%E8%BF%99%E4%B8%AA%E5%81%9A%E6%B3%95%E4%B8%BA%E4%BB%80%E4%B9%88%E6%98%AF%E9%94%99%E8%AF%AF%E7%9A%84%3F%E6%AD%A3%E7%A1%AE%E8%A7%A3%E6%B3%95%E6%98%AF%E4%BB%80%E4%B9%88%3F)
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设实数x,y,m,n满足x^2+y^2=3,m^2+n^2=1,若a≥mx+ny恒成立,求a的取值范围(x-m)^2≥0(x^2+m^2)/2≥xm(y-n)^2≥0(y^2+n^2)/2≥yn(x^2+m^2+y^2+n^2)/2≥xm+yn2≥xm+yna≥xm+yna≥2这个做法为什么是错误的?正确解法是什么?
设实数x,y,m,n满足x^2+y^2=3,m^2+n^2=1,若a≥mx+ny恒成立,求a的取值范围
(x-m)^2≥0
(x^2+m^2)/2≥xm
(y-n)^2≥0
(y^2+n^2)/2≥yn
(x^2+m^2+y^2+n^2)/2≥xm+yn
2≥xm+yn
a≥xm+yn
a≥2
这个做法为什么是错误的?
正确解法是什么?
设实数x,y,m,n满足x^2+y^2=3,m^2+n^2=1,若a≥mx+ny恒成立,求a的取值范围(x-m)^2≥0(x^2+m^2)/2≥xm(y-n)^2≥0(y^2+n^2)/2≥yn(x^2+m^2+y^2+n^2)/2≥xm+yn2≥xm+yna≥xm+yna≥2这个做法为什么是错误的?正确解法是什么?
x²+y²=3,设:x=√3cosa、y=√3sina;
m²+n²=1,设:m=cosb、n=sinb,
mx+ny
=√3cosacosb+√3sinasinb
=√3cos(a-b)
mx+ny的取值范围是:[-√3,√3]
-√3,√3
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