y=ax b分之x的平方 x-1值域

若y=根号[(x^2-2)/(5x-4)]+[根号(x^2-2)/(4-5x)]+2[(x^2-2)/(5x-4)]>=0(x^2-2)/(4-5x)]>=0两个相反数同时大于或等于0,两个必须都为0

x-3y=0x=3y(2x+y)/(x^2-2xy+y^2)÷1/(x-y)=(2x+y)/(x-y)^2÷1/(x-y)=(2x+y)/(x-y)^2*(x-y)=(2x+y)/(x-y)=(2*3

(x分之x的平方-y的平方)乘以(x的平方-2xy+y的平方分之2x)其中x=2y=1=(x^2-y^2)/x*2x/(x-y)^2=(x+y)(x-y)/x*2x/(x-y)^2=2(x+y)/(x

a×(b－c)

（x³＋3x²y-9xy²）-1/2（6x²y＋4x²）=x³＋3x²y-9xy²-3x²y-2x²

原式=（x²+2x+1）+（y²-8y+16）=（x+1）²+（y-4）²=0∴x+1=0,y-4=0∴x=-1,y=4y/x=4/-1=-4

已知5x-4y=0,切xy≠0,x/y=4/5;求(x平方-y平方)分之(x平方+y平方)-(x-y)分之(x+y)的值=(x²+y²)/(x²-y²)-(x+

x/y=1/2y=2x(x^2+2xy+y^2)/(x^2-xy+y^2)=(x^2+2x*2x+(2x)^2)/(x^2-x*2x+(2x)^2)=9x^2/3x^2=3

y分之x=2分之1y=2x所以原原式=(x²-2x²-8x²)分之（x²＋4x²＋4x²）=-9分之9=-1

题目是不是Y=x-根号（x^2-4）/2+根号（4-x^2）+1再问：Y=根号（x^2-4）/X-2后面基本相同内阁加1是在根号后面的再答：那么这题很简单的,,x^2-4≥0所以x≥2或x≤-2又4-

x+1/x=4=>x=2-3^0.5=>x^2=4x-1;x^2+1=4x=>x^4+x^2/(1+x^2)=(4x-1)^2+x^2/(4x)=16x^2-8x+1+x/4=16(4x-1)-8x+

y=[(x+1)²/(x+1)(x-1)]÷[(x+1)/x(x-1)]-x+1=(x+1)/(x-1)×x(x-1)/(x+1)-x+1=x-x+1=1所以y和x的取值无关

x+7y=y-3x4x=-6yy=-2x/3所以y²=4x²/9所以原式=(x²+4x²/9)/(x²-4x²/9)=(13x²/

3∵x/y=1/2∴y=2x故（x^2+2xy+y^2)/(x^2-xy+y^2)=(x+y)^2/(x^2-xy+y^2)=(x+2x)^2/(x^2-x·2x+4x^2)=9x^2/(3x^2)=

答案是2解.x（x-1）-(x的平方-y)=-2x的平方-x-x的平方+y=-2y-x=-2两边平方得：（y-x）的平方=4y的平方-2xy+x的平方=4两边同除以2得：（x的平方+y的平方）/2-x

（1-y+x分之y）÷y的平方-x的平方分之x=[(y+x-y)/(y+x)]÷x/(y²-x²)=x/(y+x)*[(y+x)(y-x)/x]=y-x

y分之x=2分之1y=2x所以原原式=(x²-2x²-8x²)分之（x²＋4x²＋4x²）=-9分之9=-1

x-y分之2x的平方-y-x分之x的平方-4xy+x-y分之2y的平方-x的平方=2x/(x-y)-(x²-4xy)/(y-x)+(2y-x²)/(x-y)=(2x+x²

y’=[sinx/(1-x²)]'=[(sinx)'(1-x²)-sinx(1-x²)']/(1-x²)²=[cosx(1-x²)+2xsi

x(x-1)-(x²-y)=-2x²-x-x²+y=-2-x+y=-2x-y=2(x²+y²)/2-xy=(x²-2xy+y²)/