若动点(x,y)在直线2x y-2=0上运动,则9^x 3^y的最小值为-搜答案-搜搜做题作业网

两个截距分别带入x=0得到y轴截距2y=0x1所以定义域三角形面积为1f(x,y)=1在上述给定区域fX(x)=∫(0~2-2x)1dy=2-2x0

对.前提是x不等于y

y是多少原式=(x+y)(x-y)/(x-y)²+x(y+3)/x(y+3)=(x+y)/(x-y)+1=(x+y+x-y)/(x-y)=2x/(x-y)=1/(1/2-y)再问：y=3

(x+y-2xy)(x+y-2)+(1-xy)^2=(x+y)^2-2(1+xy)(x+y)+4xy+1-2xy+x^2y^2=(x+y)^2-2(1+xy)(x+y)+(1+xy)^2=(x+y)^

=xy(x-y)×[-xy/(x-y)]=-(xy)²=-x²y²

2x方y和3x方y是同类项,剩下两个是同类项,合并后为5x方y和-3x方y

【x²+xy/（x-y）】/【xy/（x-y）】=【x²（x-y）/（x-y）+xy/（x-y）】/【xy/（x-y）】={【x²（x-y）+xy】/（x-y）}/【xy

[2x(x^2y-xy^2)+xy(xy-x^2)]÷(x^2y)=[2x*xy（x-y）+xy*x（y-x）]÷(x^2y)=[2x^2*y(x-y)-x^2*y(x-y)]÷(x^2y)=x^2*

设：改点为（x0,y0）根据距离公式算出平方和,得到一个二元函数表达式,对其求偏导数,得到稳定点,后验证黑塞矩阵的正定或者负定或者不定证明其为极小值点,证毕,得出答案,该点即为所求.（目测是正定从而取

：（xy-x²）÷xy分之x²-2xy+y²乘x²分之x-y=-x(x-y)×xy/(x-y)²×(x-y)/x²=-y∴结果与x的值没有关

不能算出AB的方程然后和双曲线联列一下方程,因为AB在同侧,算最小值你也可以做B关于X轴对称点C,然后求出AC方程,与双曲线联立.案答案来解释,可以设B与F交双曲线于P点,要求AP+BP最小,FP-A

(x+y)²-(x-y)(x+y)-(xy)²÷(xy)=x²+y²+2xy-(x²-y²)-xy=2y²+xy.方程组｛3x+2

是这样不?

2x^2y*（-xy)=-2x^3y^,3x^2y*(-xy)=-3x^3y^,-2x^3y^-3x^3y^=-5x^3y^.还有不同的解法,留给您练习.

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

原式=-x²y+2xy²+2x²y²-4xy³-3x³+27x³y-18x²y²+2x²y-18x&

先化简在求值(-1/3XY)²×[xy(2x-y)-2x(xy-y)²],=1/9x²y²*(2x²y-xy²-2x²y+2xy&

xy+y+(k-5)x+2=0(1)x-y-k=0(2)由(2)式得y=x-k,代入(1)式得：x(x-k)+(x-k)+(k-5)x+2=0x²-4x-k+2=0x=2±√(2+k),k≥