1.解方程:12-2x+xcosα=024cosα-2xcosα+x(cos^2 α-sin^2 a)=0 请详解2.解方程:b+2c+ λbc=0a+2c+ λac=02a+2b+ λab=0 请详解3,在平面xOy上求一点,使它到x=0,y=0及x+2y-16=0三直线的距离平方之和为最小.设所求点为(x
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![1.解方程:12-2x+xcosα=024cosα-2xcosα+x(cos^2 α-sin^2 a)=0 请详解2.解方程:b+2c+ λbc=0a+2c+ λac=02a+2b+ λab=0 请详解3,在平面xOy上求一点,使它到x=0,y=0及x+2y-16=0三直线的距离平方之和为最小.设所求点为(x](/uploads/image/z/1144384-16-4.jpg?t=1.%E8%A7%A3%E6%96%B9%E7%A8%8B%EF%BC%9A12-2x%2Bxcos%CE%B1%3D024cos%CE%B1-2xcos%CE%B1%2Bx%28cos%5E2+%CE%B1-sin%5E2+a%29%3D0+%E8%AF%B7%E8%AF%A6%E8%A7%A32.%E8%A7%A3%E6%96%B9%E7%A8%8B%EF%BC%9Ab%2B2c%2B+%CE%BBbc%3D0a%2B2c%2B+%CE%BBac%3D02a%2B2b%2B+%CE%BBab%3D0+%E8%AF%B7%E8%AF%A6%E8%A7%A33%2C%E5%9C%A8%E5%B9%B3%E9%9D%A2xOy%E4%B8%8A%E6%B1%82%E4%B8%80%E7%82%B9%2C%E4%BD%BF%E5%AE%83%E5%88%B0x%3D0%2Cy%3D0%E5%8F%8Ax%2B2y-16%3D0%E4%B8%89%E7%9B%B4%E7%BA%BF%E7%9A%84%E8%B7%9D%E7%A6%BB%E5%B9%B3%E6%96%B9%E4%B9%8B%E5%92%8C%E4%B8%BA%E6%9C%80%E5%B0%8F.%E8%AE%BE%E6%89%80%E6%B1%82%E7%82%B9%E4%B8%BA%EF%BC%88x)
1.解方程:12-2x+xcosα=024cosα-2xcosα+x(cos^2 α-sin^2 a)=0 请详解2.解方程:b+2c+ λbc=0a+2c+ λac=02a+2b+ λab=0 请详解3,在平面xOy上求一点,使它到x=0,y=0及x+2y-16=0三直线的距离平方之和为最小.设所求点为(x
1.解方程:12-2x+xcosα=0
24cosα-2xcosα+x(cos^2 α-sin^2 a)=0 请详解
2.解方程:b+2c+ λbc=0
a+2c+ λac=0
2a+2b+ λab=0 请详解
3,在平面xOy上求一点,使它到x=0,y=0及x+2y-16=0三直线的距离平方之和为最小.
设所求点为(x,y),则此点到三直线的距离一次为:|x|,|y|,|x+2y-16|/√5
我想知道这√5是怎么求出来的?
1.解方程:12-2x+xcosα=024cosα-2xcosα+x(cos^2 α-sin^2 a)=0 请详解2.解方程:b+2c+ λbc=0a+2c+ λac=02a+2b+ λab=0 请详解3,在平面xOy上求一点,使它到x=0,y=0及x+2y-16=0三直线的距离平方之和为最小.设所求点为(x
3.点(x0,y0)到直线Ax+By+c=0的距离公式d=|Ax0+By0+C|/√(A²+B²),
故点(x,y)到直线x+2y-16=0的距离是|x+2y-16|/√(1²+2²)=|x+2y-16|/√5.
∵x^2+xcosαcosβ+cosγ-1=0的两个根为x1、x2,
∴x1+x2=-cosαcosβ,x1x2=cosγ-1
又x1+x2=x1x2
∴-cosαcosβ=cosγ-1
cosαcosβ+cosγ=1
又∵α、β、γ是三角形内角
∴γ=180°-(α+β)
∴cosγ=cos[180°-(α+β)]
...
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∵x^2+xcosαcosβ+cosγ-1=0的两个根为x1、x2,
∴x1+x2=-cosαcosβ,x1x2=cosγ-1
又x1+x2=x1x2
∴-cosαcosβ=cosγ-1
cosαcosβ+cosγ=1
又∵α、β、γ是三角形内角
∴γ=180°-(α+β)
∴cosγ=cos[180°-(α+β)]
=-cos(α+β)
=-cosαcosβ+sinαsinβ
∴cosαcosβ-cosαcosβ+sinαsinβ=1
sinαsinβ=1
sinα=1/sinβ
又0
∴sinα=sinβ=1
α=β=π/2
∴无法判断
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