求圆(x-1)² (y-2)²=1关于直线y=x

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求下列函数的值域: (1)y=1-x²/1+x² (2)y=-x²-2x+3 (3)y=x+1/x (4)y=x+√1-

解题思路:用x2的取值范围、二次函数的的性质、均值不等式,换元法求函数的值域解题过程:

当x+y/x-y=1/2时,求代数式x-y/x+y-2x+2y/x-y的值

x+y/x-y=1/2取倒数x-y/x+y=2所以x-y/x+y-2x+2y/x-y=x-y/x+y-2(x+y/x-y)=2-2×1/2=2-1=1

2x^2+y^2=√2 2x+y=1 求x,y

由2x+y=1得y=1-2x代入2X^2+y^2=2(或根号二)得2x^2+(1-2X)^2=2(或根号二)6X^2-4x+1=2(或根号二)直接用求根公式即可

已知p(x y) 在圆x^2+y^2=1上 求y-2x的取取值范畴

令x=cosa则y²=1-cos²a=sin²a所以y=sina所以y-2x=sina-2cosa=√(1²+2²)sin(a-b)=√5sin(a-

1/x-1/y=3,求2x+3xy-2y/x-y

∵1/x-1/y=3∴y-x=3xy2x+3xy-2y/x-y=2(x-y)+3xy/x-y=-6xy+3xy/-3xy=1

已知:3x=8y.求(1)x+y/y (2)2x+3y/x-2y

3x=8yx/y=8/3(1)x+y/y=x/y+1=8/3+1=11/3(2)2x+3y/x-2y分子分母同时除以y得=(2x/y+3)/(x/y-2)=(16/3+3)/(8/3-2)=(25/3

x+y=1,xy=-1/2,求x(x+y)(x-y)-x(x+y)2

x(x+y)(x-y)-x(x+y)2=x(x+y)[(x-y)-(x+y)]=x(x+y)(-2y)=-2xy(x+y)=-2×(-1/2)×1=1再问:18p3q3-2pq再答:7(x-1)3-1

圆x^2+y^2=1,求点[(x(x+y),y(x+y)]的轨迹方程

令m=x(x+y),n=y(x+y)则m-n=x^2-y^2,于是m-n+1=2x^2,1-m+n=2y^2(m+n-1)^2=4x^2y^2=(m-n+1)(1-m+n)整理得:m^2+n^2-m-

若|x+y-1|+(x-y-2)²=0,求代数式(x+2y)(x-2y)-(2x-y)(-y-2x)的值.

x+y=1x-y=2(x+2y)(x-2y)-(2x-y)(-y-2x)=(x+2y)(x-2y)+(2x-y)(y+2x)=x²-4y²+4x²-y²=5x&

2^x=4^(y-1) 27^y=3^(x+1)求x-y

把4看成2^22^x=2^2(y-1)得出x=2(y-1)27^y=3^3y=3^(x+1)得出3y=(x+1)联立两个方程就可以解决啦y=-1x=-4x-y=-3

已知x+y=0,x+13y=1,求x²+12xy+13y²的值.

解题思路::∵x+y=0,x+13y=1,解得x=1/12,y=-1/12∴x²+12xy+13y²=1/144-1/12+13/144=14/144-1/12=2/144=1/72解题过程:已知x+

求y‘-(1/x)y=x^2 的通解

即xy'-y=x^3即(xy'-x'y)/x^2=x即(y/x)'=xy/x=1/2x^2+cy=x(1/2x^2+c);c为常数

已知4x=9y求(1)x+y/y (2)y-x/2x

4x=9yx=9/4*y(1)(x+y)/y=[(9/4)y+y]/y=(9/4+1)y/y=9/4+1=13/4(2)(y-x)/2x=[y-(9/4)y]/[2*(9/4)y]=(1-9/4)y/

x+3y=3,求x²-9y分之2x-6y的值

解题思路:本题目应该是9y的平方吧,要不然没法做出来。解题过程:(2x-6y)/(x的平方-9y)=2(x-3y)/[(x-3y)(x+3y)]=2/(x+3y)=2/3

y=ln(x+√x^2+1),求y

x≤0时√x^2=-x所以y=0x>0时√x^2=x所以y=ln(2x+1)

y=3x²-x+2,x∈[1,3],求值域,(求详细解题步骤)

解题思路:利用二次函数的图像的对称轴进行分析即可得了,解题过程:见附件

已知x²+y²+5=2x+4y,求【2x²-(x-y)(x-y)】【(x+y-1)(x-y

1,-3再问:过程。。。再答:★(x²-2x)+(y²-4y)=5★(x-1)²+(y-2)²=1+4-5★(x-l)²=0,(y-2)²=

若|x+2y-1|+y²+4y+4=0,求(2x-y)²-2(2x-y)(x+2y)+(x+2y)&

∵|x+2y-1|+y²+4y+4=0∴|x+2y-1|+(y+2)²=0∴x=5,y=-2(2x-y)²-2(2x-y)(x+2y)+(x+2y)²=[(2x

1奥数题x-2y/x+2y=3,求x-2y/3(x+2y)-3(x-2y)/x-2y的差

(x-2y)/(x+2y)=3取倒数(x+2y)/(x-2y)=1/3所以原式=(1/3)[(x-2y)/(x+2y)]-3[(x+2y)/(x-2y)]=(1/3)×3-3×(1/3)=0

已知x=1/3,y=-1/2,求代数式x-(x+y)+(x+2y)-(x+3y)+(x+4y)-(x+5y)+...-(

原式=x-x+x-x+……-x+(2-1+4-3+5-4+……+2008-2007-2009)y=0+(1×1004-2009)y=-1005y=1005/2