y=kx ksqrt(10-x)的最大值

1.x=1y=1.72.2983.248189

(4x-2y-z)-{5x[8y-2y-(x+y)]-x+(3y-10z)]=4x-2y-z-5x[6y-(x+y)]+x-(3y-10z)=4x-2y-z-30xy+5x²+5xy+x-3

x-y/7-x+y/10=1/2两边乘7010x-10y-7x-7y=353x-17y=35(1)2(x-y)+5(x+y)=32x-2y+5x+5y=37x+3y=3(2)(2)*3-(1)*79y

(x+y)/6+(x-y)/10=3①(x+y)/6+(y-x)/10=-1②①可变为：(x+y)/6-（y-x)/10=3③②+③=(x+y)/3=2得x+y=6④②-③=（y-x）/5=-4得y-

(7x-10y)-(-10y-5x)=7x-10y+10y+5x=12x再问：7x-10y+10y+5x似乎等于2x+10y再答：-10y+10y=0

先化简[(x²+y²)-(x-y)²+2y(x-y)]/4y[(x²+y²)-(x-y)²+2y(x-y)]/4y=[x²+y&s

x-y-3绝对值+(2x+2y+10)的平方=0X-Y-3=0X-Y=32X+2Y+10=0X+Y=-5(x+y)(x-Y)=3X(-5）=-15

|x|+x+y=10|y|+x-y=12两式相加得|x|+|y|+2x=22.(1)两式相减得|x|-|y|+2y=-2.(2)所以-|y|+2y＜0若y＞0,则显然-y+2y＜0,即y＜0,矛盾若y

若是y的10次,我也不必说了若是y的10阶导数1：1/（1+x）2：-（1+x）^(-2)3：2（1+x）^(-3)……10：9!（1+x）^(-9)

在坐标系中先作2x+5y≥10对应的直线2x+5y=10,取直线上方区域；然后作2x-3y≥-6对应的直线2x-3y=-6,把y的系数变为正,可知取直线下方区域；最后作2x+y≤10对应的直线2x+y

.死算呗xy-2x+3y-6=xy+10x-3y-30,2x-y=4xy+3x+y+3=xy+12x-2y-24,3x-y=92式减1式,x=5,y=6

(100X+Y)-(10Y+X)=(10Y+X)-(10X+Y)100X+Y-10Y-X=10Y+X-10X-Y99X-9Y=9Y-9X-18Y=-108XY=6X把Y=6X代入X+Y=7得X+6X=

2/(x+y)+5/(x-y)=189/(x+y)-10/(y-x)=66设x+y=a,x-y=b,原方程化为2/a+5/b=18-----1(1)9/a+10/b=66-------(2)(2)-(

已知2x-y=10,式子[(x^2+y^2)-(x-y)^2+2y(x-y)]/4y=[x^2+y^2-x^2+2xy-y^2+2xy-2y^2]/4y=[4xy-2y^2]/4y=2y(2x-y)/

另a=1/（x+y）,b=1/(x-y);则2a+5b=18；9a+10b=66；解得a=6,b=6/5；即       &nbs

①－②可得：x-y/10－y-x/10=4x-y／5＝4x-y＝20——③①＋②可得：x+y/3=2x+y=6——④由③④可得：x=13,y=－7

(x+y)(x+y-10)+25=0设x+y=aa(a-10)+25=0a²-10a+25=0(a-5)²=0a=5∴x+y=5x-y=3x=4y=1

（1）x^2+y^2-4x+10y+29=0x^2-4x+4+y^2+10y+25=0(x-2)^2+(y+5)^2=0x=2y=-5剩下的自己解下面两个题目是不是写错了,没看懂

若x≤0,|x|=-x|x|+x+y=10y=10代入x+|y|-y=12得x=12>0矛盾,∴x>02x+y=10①若y≥0,x+|y|-y=x=12y=10-2x∴yx-2y=12②联立①②解得x

y=6-x,9x