已知实数x y满足|x-5| √y 4=0,求式子(x y)∨2016
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将其看出关于x的方程5x²-(6y+4)x+2y²+2y+1=0其判别式△≥0而△=(6y+4)²-20(2y²+2y+1)=-4y²+8y-4=-4
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=1/10(2+3y-Sqrt[-1+2y-y^2],x=1/10(2+3y+Sqrt[-1+2y-y^2])
因为2x^2+4xy+5y^2-4x+2y+5=0所以(x^2+4xy+4y^2)+(x^2-4x+4)+(y^2+2y+1)=0即(x+y)^2+(x-2)^2+(y+1)^2=0也即x+2y=0,
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
5x^2-3xy+1/2y^2-2x+1/2y+1/4=0乘以4通分之后得到:20x^2-12xy+2y^2-8x+2y+1=0(4x-y)^2+(2x-y)^2-2(4x-y)+1=0(4x-y-1
|x+y+7|+(xy-5)^2=0,则x+y+7=0,x+y=-7,xy-5=0,xy=5,所以3x^2y^3+3x^3y^2=3x^2y^2(y+x)=3*(xy)^2*(x+y)=3*5^2*(
两边除以y得x/y-5√(x/y)-6=0∴[√(x/y)-6][√(x/y)+1]=0∴√(x/y)=6√(x/y)=-1(舍去)∴x/y=36
⑴若x=y,则x、y是方程m^2+√2m=√3的两个相等实根由根与系数关系得:x+y=-√2,xy=-√3⑵若x≠y,两式相减得:x^2-y^2+√2y-√2x=0,(x+y-√2)(x-y)=0得:
x>=4x/y=x-yx=(x-y)yx=xy-y2y2=x(y-1)x=y2/(y-1)设y-1=t因为y>1所以t>0故x=(t2+2t+1)/tx=t+1/t+2>=2根号1+2x>=4
|x+y-7|+√(xy-6)=0x+y-7=0xy=6x=1,y=6,或x=6,y=1x=1,y=6时(x+2y)÷(y-x)=13÷5=13/5x=6,y=1时(x+2y)÷(y-x)=8÷(-5
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
y=√(2x-1)+√(1-2x)+2因为被开方数要大于等于0所以2x-1≥0且1-2x≥0x≥1/2且x≤1/2所以x=1/2y=0+0+2=2xy=(1/2)×2=1因为(±1)²=1所
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
y=-x²+x+3x+y=-x²+2x+3=-x²+2x-1+4=-(x-1)²+4因为-1<0所以当x=1时,x+y的最大值=4
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
设k=9x-y,则y=9x-k,代入已知式,得-4≤x-(9x-k)≤-1,-1≤4-(9x-k)≤5(改题了),即8x-4≤k≤8x-1,9x-5≤k≤9x+1,画示意图知,由k=8x-4,k=9x
由x2+xy+y2=3得,x^2+y^2=3-xyx^2+y^2≥2xy得,xy≤1所以x^2-xy+y^2=3-2xy≥1等号成立当且仅当x=y=±1