若x、y满足条件x²-y²=1,那么1/x²+2y/x的取值范围是
来源:学生作业帮 编辑:神马作文网作业帮 分类:数学作业 时间:2024/11/13 18:05:39
若x、y满足条件x²-y²=1,那么1/x²+2y/x的取值范围是
x²-y²=1
1/x²+2y/x
设 x=secm,
则 y=tanm
1/x²+2y/x
=cos^m+2tanm*cosm
=cos^m+2sinm
=1-sin^m+2sinm
=2-(sinm-1)^2
所以
-2
再问: 谢谢,知道了。 那请问cos²36°+cos²72°=?谢谢、
再答: cos²36°+cos²72° =cos²36°+1/2(2sin^18°-1+1) =cos²36°+1/2(-cos36°+1) =cos²36°-1/2cos36°+1/2
再问: 答案是3/4
再答: 原式=1-sin²36+sin²18 =1+(sin18+sin36)(sin18-sin36) =1+[sin(27-9)+sin(27+9)][sin(27-9)-sin(27+9)] =1+(sin27cos9-cos27sin9+sin27cos9+cos27sin9)(sin27cos9-cos27sin9-sin27cos9-cos27sin9) =1-4sin27cos9cos27sin9 =1-(2sin27cos27)(2sin9cos9) =1-sin18sin54 =1-2sin18cos18cos36/2cos18 =1-sin36cos36/2cos18 =1-sin72/4sin72 =1-1/4 =3/4
1/x²+2y/x
设 x=secm,
则 y=tanm
1/x²+2y/x
=cos^m+2tanm*cosm
=cos^m+2sinm
=1-sin^m+2sinm
=2-(sinm-1)^2
所以
-2
再问: 谢谢,知道了。 那请问cos²36°+cos²72°=?谢谢、
再答: cos²36°+cos²72° =cos²36°+1/2(2sin^18°-1+1) =cos²36°+1/2(-cos36°+1) =cos²36°-1/2cos36°+1/2
再问: 答案是3/4
再答: 原式=1-sin²36+sin²18 =1+(sin18+sin36)(sin18-sin36) =1+[sin(27-9)+sin(27+9)][sin(27-9)-sin(27+9)] =1+(sin27cos9-cos27sin9+sin27cos9+cos27sin9)(sin27cos9-cos27sin9-sin27cos9-cos27sin9) =1-4sin27cos9cos27sin9 =1-(2sin27cos27)(2sin9cos9) =1-sin18sin54 =1-2sin18cos18cos36/2cos18 =1-sin36cos36/2cos18 =1-sin72/4sin72 =1-1/4 =3/4
若实数x、y满足条件x²-y²=1,则(1/x²)+(2y/x)的取值范围是?
若实数x,y满足2x²+3y²=1,S=3x²-2y²,则S的取值范围是
实数x,y满足x+y≤2,(x-1)²+(y-1)²≤2,则2x+y的取值范围是
实数X.Y满足X*X+Y*Y-2X-2Y+1=0,则(Y-4)/(X-2)的取值范围是?
若实数x,y满足2x*X+3y*y=1,S=3x*x-2y*y,则S的取值范围为
若实数X Y满足y=2x+1,且x小于等于2y大于等于3,则y/x的取值范围是 ?
方程组中:x+2y=1+m ,2(x+1)>14-x,若未知数x,y 满足x+y>0,则m的取值范围是?
已知实数x,y满足x²+y²-2y=0,(1)求2x+y的取值范围
已知实数X,Y满足X²+Y²+4X+3=0,求Y-2/X-1的取值范围.
若实数x,y满足{x-y+1≥,x>0},则y/(x-1)的取值范围是
若对满足条件x+y+3=xy(x>0,y>0)的任意x,y,(x+y)2-a(x+y)+1≥0恒成立,则实数a的取值范围
已知实数x,y满足x²+4y²=4x,求x+y的取值范围