已知实数xy满足y=8-2x分之

实数x,y,z,满足那么x+y=6,z^2=xy-9,∴xy=z^+9,(x-y)^=(x+y)^-4xy=-4z^>=0,∴z=0,(x+y)^z=6^0=1.

再问：该方法此处计算是错的，应该为，接下来的都不对了再答：那就从那步开始吧x+y=xy-8若x，y大于0xy-8=x+y≥2√xyxy-8≥2√xyxy-2√xy-8≥0(√xy-4)(√xy+2)≥

x^2+y^2-6x+4y+13=0x^2-6x+9+y^2+4x+4=0(x-3)^2+(y+2)^2=0平方项恒非负,两平方项之和等于0,则两平方项都等于0.x-3=0x=3y+2=0y=-2x^

x²+y²-xy+2x-y+1=0x²+2x+1-y(x+1)+y²=0(x+1)²-y(x+1)+y²=0(x+1-y/2)²+

由已知x,y正实数由2x+2y+xy=5得5-xy=2(x+y)≧2*2√(xy)所以xy+4√(xy)-5≤0[√(xy)+5][√(xy)-1]≤00＜√(xy)≤1故,0

答：x>0,y>0x-√(xy)-2y=0(√x-2√y)(√x+√y)=0因为：x>0,y>0所以：√x+√y>0所以：√x-2√y=0所以：√x=2√y所以：x=4y所以：[x+3√(xy)+2y

x^2+2xy+y^2-(x+y)-6=0(x+y)^2-(x+y)-6=0令x+y为a即a^2-a-6=0(a-3)(a+2)=0所以a=3或a=-2故x+y=3或-2

x^2+2y^2+2x+2=2xy(x-y)^2+y^2+2x+2=0(x-y)^2+(y+1)^2+2x-2y+1=0(x-y)^2+2(x-y)+1+(y+1)^2=0[(x-y)+1]^2+(y

解由题知求xy的最大值,则x,y必定同号,不妨设x,y同正则由x^2+y^2+xy=1/3得1/3=xy+x²+y²即1/3-xy=x²+y²≥2xy即1/3≥

令x=sinay=cosa(1-xy)(1+xy)=1-（xy）^2=1-(sinacosa)^2=1-1/4sin(2a)^2显然0《（sin2a）^2《13/4《1-1/4sin(2a)^2《1即

x^2+xy+y^2=3设t=x+2yx=t-2y(t-2y)^2+(t-2Y)y+y^2=3t^2-4yt+4y^2+ty-2y^2+y^2=33y^2-(4t-t)y+t^2-3=03y^2-(3

2X^2+4XY+4Y^2+8X+12Y+10=2(x+y)^2+2Y^2+8x+12y+10=2(x+y)^2+8(x+y)+8+[2Y^2+4y+2]=2[(x+y)+2]^2+2(y+1)^2=

由2x^2+4xy+4y^2+8x+12y+10=0得x^2+2xy+2y^2+4x+6y+5=0x^2+2(y+2)x+(y+2)^2-(y+2)^2+2y^2+6y+5=0(x+y+2)^2+(y

分解因式有(x-3y)(2x-y)=0所以有x=3y或2x=y所以x:y=3:1或x:y=1:2

z=3x+y=13(x+2y)/6+5(x-4y)/6当x=5,y=2时取到,z最大值17

x²+y²-xy+2x-y+1=[3（x+1）²+(x-2y+1）²]/4=0,由于（x+1）²>=0且(x-2y+1）²>=0,则有x+1

(x+1)^2+(y+2)^2=25(x,y)在以（-1,-2）为圆心,5为半径的圆上用线性规划思想,（1）设k=y/x,则只需y=kx与圆有公共点即可,用圆心到直线距离=0,k为全体实数（2）x^2

∵正实数x、y满足x+2y=xy，∴1y+2x=1（x＞0，y＞0），∴2x+y=（2x+y）•1=（2x+y）•（1y+2x）=2xy+2yx+1+4≥22xy•2yx+5=9（当且仅当x=y=3时

10x²-2xy+y²+6x+1=0（3x+1）²＋(x-y)²=03x+1=0x-y=0所以x=y=-1/3x+y=-2/3再问：3x+1=x-y=再答：3x

x^2+xy-y^2=0(同时除以y^2)(x/y)^2+x/y-1=0用解二元一次方程的方法得x/y=(-1±√5)/2