x² 2y² 2xy-4y 4=0求x的y平方
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原式分解因式得x^3y^3(2x-y)=(xy)^3(2x-y)=8/3.(x^3表示x的3次方)
把原式两边对x求导得:x^2+12y^3*dy/dx+1+2dy/dx=0合并同类项移项得:dy/dx=-(1+2x)/(12y^3+2)
X²+Y²+X²Y²-4XY+1=0(X-Y)^2+(XY-1)^2=0所以X-Y=0XY-1=0(X-Y)^2008-(XY)^2008=0^2008-1^2
原方程组可化为4x-3y=12 ①3x-4y=2 ②,①×4-②×3,得7x=42,解得x=6.把x=6代入①,得y=4.所以方程组的解为x=6y=4.
∵x+y=6,xy=4,∴(1)x2+y2=(x+y)2-2xy,=62-2×4,=28;(2)(x-y)2=x2+y2-2xy,=28-2×4,=20;(3)x4+y4=(x2+y2)2-2x2y2
(x⁴+y⁴)÷(xy)²=[(x²+y²)²-2x²y²]/(x²y²)=[(4xy)
(x+y+z)^2=[(x+y)+z]^2=(x^2+2xy+y^2)+z^2+2zx+2zy=x^2+y^2+z^2+2xy+2xz+2yz=x^2+y^2+z^2+2(xy+xz+yz)=0x+y
3x^2+xy-2y^2=0推出(3x-2y)(x+y)=0推出x=-y或x=(2/3)y{(x+y)/(x-y)+4xy/(y^2-x^2)}/{(x+3y)*(x-y)}/x^2-9y^2推出:{
即(x-2y)²=0x-2y=0所以x=2y所以原式=(2x²+2xy-xy-y²)/(4x²-4xy+y²)=(2x²+xy-y²
x2+y2=(x+y)2-2xy=14x3+y3=(x2+y2)×(x+y)-xy2-yx2=14×4-xy(x+y)=52……剩下的就是这么个算法,手机党,求个最佳哈
方程ax^2+bx+c=0,判断这个方程有没有实数根,有几个实数根,就要用ΔΔ=b^2-4ac若Δ<0,则方程没有实数根Δ=0,则方程有两个相等实数根,也即只有一个实数根Δ>0,则方程有两个不相等的实
(x2+z2)(x2+y2)(y2+z2)=(x+y)2-2xy×(x+z)2-2xz×(y+z)2-2yz--之后不清楚了
2x+y+3=xy,①若x=1,则y不存在,此时5x+4y无最值;②若x≠1,则y=(2x+3)/(x-1),M=5x+4y=5x+(8x+12)/(x-1)=(5x²+3x+12)/(x-
x+y=5xy(2x-3xy+2y)/(x+2xy+y)=[2(x+y)-3xy]/[(x+y)+2xy]=(2×5xy-3xy)/(5xy+2xy)=7xy/7xy=1再问:若x+1/x=3,求(x
2x+y=0,求4x²+2xy(x+y)+y³→4x²应改为4x³,即:4x³+2xy(x+y)+y³4x³+2xy(x+y)+y
∵x-y=4xy∴原式=[2(x-y)+3xy]/[-(x-y)-2xy]=(8xy+3xy)/(-4xy-2xy)=11xy/(-6xy)=-11/6
(x^4-y^4)÷(x^2+y^2)/(x+y)=(x^2-y^2)(x^2+y^2)÷(x^2+y^2)/(x+y)=(x^2-y^2)(x^2+y^2)*(x+y)/(x^2+y^2)=(x^2
0=x^2+y^2+x^2y^2-4xy+1=x^2+y^2-2xy+x^2y^2-2xy+1=(x-y)^2+(xy-1)^2,两个平方数的和等于0,所以,x=y,xy=1,带入得(x-y)^200
因为x²+4y²+x²y²-6xy+1=0(x²-4xy+4y²)+(x²y²-2xy+1)=0(x-2y)²