x y)^2=7,(x-y)^2=4,则xy
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已知x/y=7/2,(2x²+3xy+2y²)/(x²+2xy+y²)=(2x/y+3+2y/x)/(x/y+2+y/x)=(2×7/2+3+2×2/7)/(7
(x^2-3XY+2Y^2)/(X^2-2XY+Y^2)=(x-2y)(x-y)/(x-y)^2=(x-2y)/(x-y)=(x-y-y)/(x-y)=1-y/(x-y)=1-1/(x/y-1)=1-
第一题设x+y=axy=b则x^2+y^2=a^2-2b原方程变为a^2-2b+b=13①b+a=7②这就变成了简单的二元二次方程由②得b=7-a带入①得a^2+a-20=0(a-4)(a+5)=0得
对.前提是x不等于y
(-3x^y+2xy)-(4x^+xy)=-3x^y+2xy-4x^-xy=-3x^y+xy-4x^所以填上-3x^y+xy-4x^
原式=6x+4xy+2xy+3x+2y-4xy+7y=9x+9y+2xy=9(x+y)+2xy=9×4+2×2=36+4=40
x^3y(-4y)^2+(-7xy)^2*(-xy)-5xy^3*(-3x)^2=16x^3y^3-49x^3y^3-45^3y^3=-78x^3y^3如果本题有什么不明白可以追问,
将1/x-1/y=5通分,得y-x/xy=5,两边同时乘以xy,得y-x=5xy,带入-x+xy+y/2x+7xy-2y,得5xy+xy/-2*5xy+7xy,合并同类项,得6xy/-3xy=-2即,
x^2y+xy^2=xy(x+y)=7x6=42再问:��(x+1)^2=13��x^2+2x+3=再答:x^2+2x+3=(x+1)^2+2=13+2=15
(4xy+12y)+[7x-(3xy+4y-x)]=4xy+12y+7x-3xy-4y+x=xy+8x+8y=xy+8(x+y)=(-2)+8*3=-2+24=22
最简单的方法就是用特殊值,令x=-2,y=1然后代入所求表达式,求出其值.为-8;另外此内题的另一个解法是变化所求表达式,使它变成关于xy^2的表达式.
解题方法:分子、分母同时除一个xy(y+xy-x)/(2x+7xy-2y)=[(y+xy-x)/xy]/[(2x+7xy-2y)/xy]=(1/x+1-1/y)/(2/y+7-2/x)=[(1/x-1
(y+xy-x)/(2x+7xy-2y)(分子、分母同除一个xy)=(1/x+1-1/y)/(2/y+7-2/x)=6/(-10+7)=-2
设X=kY.kY^2/(3k+2)Y=1/8kY^2/((2k+3)Y=1/7由上式得:3k+2=8kY,2k+3=7kY所以2/(8Y-3)=3/(7Y-2)所以Y=1/2,X=1
(5xy+4x+7y)+(6x-3xy)-(4xy-3y)=5xy+4x+7y+6x-3xy-4xy+3y=(5xy-3xy-4xy)+(4x+6x)+(7y+3y)=-2xy+10x+10y=-2x
X²+2xy+y²/xy乘x²-2xy+y²/xy+y²=(x+y)²/xy×(x-y)²/y(x+y)=(x+y)(x-y)
(-x+xy+y)/(2x+7xy-2y)=(1/x-1/y+1)/(7-2(1/x-1/y))=(5+1)/(7-10)=6/(-3)=-2
因为x-y=-2xy所以将x-y都换成-2xy了
x²-7xy+12y²=0(x-3y)(x-4y)=0x1=3yx2=4yx=3y时原式=9y²-3y²+y²/6y²=7/6x=4y时原式
[(x-y)^2+2xy-y^2]/[x^2+7xy+y^2]=[x²-2xy+y²+2xy-y²]/[x^2+7xy+y^2]=x²/[x^2+7xy+y^2