1=x² 4y²-2xy,则x 2y的取值范围
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解题思路:由完全平方公式可求解题过程:varSWOC={};SWOC.tip=false;try{SWOCX2.OpenFile("http://dayi.prcedu.com/include/rea
∵x2+y2-2x-4y+5=0,∴x2-2x+1+y2-4y+4=0,∴(x-1)2+(y-2)2=0,∴x=1,y=2,∴原式=11×2+12×3+…+12011×2012=1-12+12-13+
x-2y=12,两边平方得,x2+4y2-4xy=144,移项得,x2+4y2-1=4xy+143,将xy=-25代入上式得,x2+4y2-1=43.答:x2+4y2-1的值是:43.
已知2x-3*根号(xy)-2y=0(x>0),则x2+4xy-16y2除以2x2+xy-9y2的值是多少?2x-3*根号(xy)-2y=0(根号X-2根号Y)(2根号X+根号Y)=0根号X-2根号Y
x2+y2+3xy=(x+y)2+xy,∵x+y=4,xy=2,∴x2+y2+3xy=42+2=18.故答案为18.
设:S=x-2y,x=S+2y代入x²+y²=1中得:(s+2y)²+y²=15y²+4sy+s²-1=0∵y是实数,∴△≥0(4s)&su
∵2x2-xy-5x+y+4=0∴x2+x2-xy-4x-x+y+4=0∴x2-4x+4+x(x-y)-(x-y)=0∴(x-2)2+(x-y)(x-1)=0∵(x-2)2≥0,x≥y≥1,∴(x-y
∵4x2+y2+xy=1∴(2x+y)2-3xy=1令t=2x+y则y=t-2x∴t2-3(t-2x)x=1即6x2-3tx+t2-1=0∴△=9t2-24(t2-1)=-15t2+24≥0解得−21
∵x<y<0,∴x-y<0,x+y<0.∴x2−2xy+y2=(x−y)2=|x-y|=y-x.x2+2xy+y2=(x+y)2=|x+y|=-x-y.∴x2−2xy+y2+x2+2xy+y2=-2x
x^2+y^2+2x-4y+1=0(x+1)^2+(y-2)^2=2^2x=-1+2cosx,y=2+2sinxy-x=3+2(sinx-cosx)=3+2√2(sinx/√2-cosx/√2)=3+
x(x-1)-(x2-y)=-3x2-x-x2+y=-3-x+y=-3所以x-y=3所以原式=(x-y)2=9
解题思路:完全平方公式解题过程:varSWOC={};SWOC.tip=false;try{SWOCX2.OpenFile("http://dayi.prcedu.com/include/readq.
∵x+y=4,xy=2,∴x2+y2+3xy,=(x+y)2+xy,=42+2,=18.
由题意得(x-2)平方+(y-2)平方+(x-y)平方=0,故x=y=2,故x平方y=8
2x²+xy-3y²+x+4y-1=2×x×x+x×y-3×y×y+x+4×y-1=x×2x+x×y-y×3y+x×1+y×4-1=x×(2x+y+1)-y×(3y+4)-1再问:
x^2+y^2=5;x^2-2xy+y^2=1;2/xy/[(1/x)-(1/y)]=-2;最后得到值为1
您好:x(x-1)+(y-x2)=6x²-x+y-x²=6y-x=6x2+y2/2-xy=1/2(x²+y²-2xy)=1/2(x-y)²=1/2*6
原式=2x2-2xy-6x2+9xy-2x2+2x2-xy+y2=-2x2+5xy+2y2,当x=-1,y=-2时,原式=-2+10+8=16.
因为x²+4y²+x²y²-6xy+1=0(x²-4xy+4y²)+(x²y²-2xy+1)=0(x-2y)²