P(x0,y0)是圆x2+(y-1)2=1上一点,求x0+y0+c≥0中c的范围
问题描述：
P(x0,y0)是圆x2+(y-1)2=1上一点,求x0+y0+c≥0中c的范围
答
3x^3+2x-5
=(3x^3-3x^2)+(3x^2+2x-5)
=3x^2(x-1)+(x-1)(3x+5)
=(x-1)(3x^2+3x+5) 3x^3+2x-5
=(3x^3-3x^2)+(3x^2+2x-5)
=3x^2(x-1)+(x-1)(3x+5)
=(x-1)(3x^2+3x+5) 3x^3+2x-5
=(3x^3-3x^2)+(3x^2+2x-5)
=3x^2(x-1)+(x-1)(3x+5)
=(x-1)(3x^2+3x+5) 3x^3+2x-5
=(3x^3-3x^2)+(3x^2+2x-5)
=3x^2(x-1)+(x-1)(3x+5)
=(x-1)(3x^2+3x+5) 3x^3+2x-5
=(3x^3-3x^2)+(3x^2+2x-5)
=3x^2(x-1)+(x-1)(3x+5)
=(x-1)(3x^2+3x+5) 3x^3+2x-5
=(3x^3-3x^2)+(3x^2+2x-5)
=3x^2(x-1)+(x-1)(3x+5)
=(x-1)(3x^2+3x+5)
答
x²+(y-1)²=1令x=cosa则(y-1)²=1-cos²a=sin²ay-1=sinay=sina+1所以x+y=sina+cosa+1=√2(sina*√2/2+cosa*√2/2)+1=√2(sinacosπ/4+cosasinπ/4)+1=√2sin(a+π/4)+1所以x0+y0最小值=-√2+1...