xy-x2=1,xy2 ax2 bx a=0

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xy-x2=1,xy2 ax2 bx a=0
x(x-1)-(x2-y)=-3,求x2-y2-2xy的值

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

设x,y为实数,且x2+xy+y2=1,求x2-xy+y2的值的范围

设x2-xy+y2=Ax2-xy+y2=A与x2+xy+y2=1相加可以得到:2(x2+y2)=1+A(1)x2-xy+y2=A与x2+xy+y2=1相减得到:2xy=1-A(2)(1)+(2)×2得

求高数中微分方程通解 (1+x2)dy=(1+xy)dx

观察知,y=x是方程的特解为求通解,令y=x+t,代入原方程得(1+x^2)(1+t')dx=(1+x^2+xt)dx化简得dt/t=xdx/(1+x^2)所以,t=C(1+x^2)^(1/2)所以,

因式分解X2(X+1)-Y(XY+X)=

X²(X+1)-Y(XY+X)=X^3+X²-XY²-XY=X^3-XY²+X²-XY=X(X²-Y²)+X(X-Y)=X(X-Y

先化简,再求值:(1)2x2-5x x2 4x,其中x=-3 (2)(3x2-xy-2y2)-2(x2 xy-2y2),

:(1)2x2-5x+x2+4x,其中x=-3=3x²-x=3x(-3)²+3=27+3=30(2)(3x2-xy-2y2)-2(x2+xy-2y2),其中x=6,y=-1=3x&

x2+2y2-2xy-2y+1=0

X^2+2Y^2-2XY-2Y+1=0X^2-2XY+Y^2+Y^2-2Y+1=0(X-Y)^2+(Y-1)^2=0因为两个数的平方为一个非负数,所以得:X-Y=0Y-1=0所以:X=1Y=1所以:X

二元二次方程解法如x2+xy+y=1x2+xy+x=6

x2+xy+y=1x2+xy+x=6两个相减得x-y=5x=5+y代入任何一个式子,可以得y=-2,-6x=3,-1

因式分解(x2+x+1)(x2+x+2)=?4abx2-2(a2+b2)xy+aby2=?

(1)已经分解了,不能再分了(2)(2bx-ay)(2ax-by)

已知代数式A=2x2+3xy+2y-1,B=x2−xy+x−12

(1)A-2B=2x2+3xy+2y-1-2(x2−xy+x−12)=2x2+3xy+2y-1-2x2+2xy-2x+1=5xy+2y-2x,当x=y=-2时,A-2B=5xy+2y-2x=5×(-2

因式分解:1-x2+4xy-4y2=______.

原式=1-(x2-4xy+4y2)=1-(x-2y)2=(1+x-2y)(1-x+2y).故答案为(1+x-2y)(1-x+2y).

x2-y2=xy,xy不等于0求x2/y2=y2/x2

是求x2/y2+y2/x2=吗x2-y2=xy则x/y-y/x=1两边平方得x^2/y^2-2+y^2/x^2=1所以x^2/y^2+y^2/x^2=3

x(x-1)-(x2-y)=-3,求x2+y2-2xy的值

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

x2+y2-xy=1,则u=x2-y2的取值范围是

(x+y)^2=1+3xy(x-y)^2=1-xyu=(x+y)(x-y)|u|=√(x+y)^2√(x-y)^2=√(1+3xy)√(1-xy)=√[-3(t-1/3)^2+2/3]≤√6/3故-√

已知x(x-1)-(x2-y)=-3,求x2+y2-2xy的值

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.

已知x2+xy=5,xy+y2=-1,则x2-y2=______.

∵x2+xy=5,xy+y2=-1,∴(x2+xy)-(xy+y2)=x2+xy-xy-y2=x2-y2=5-(-1)=6.故填:6

解方程组x2+xy=12 xy+y2=4

x^2+xy=12xy+y^2=4因式分解下,得x(x+y)=12.y(x+y)=4两个方程相加,得(x+y)^2=16所以x+y=±4当x+y=4时,代入x(x+y)=12.y(x+y)=4解得x=

已知代数式:A=2x2+3xy+2y-1,B=x2-xy+x-12;

(1)∵A=2x2+3xy+2y-1,B=x2-xy+x-12,x-y=-1,xy=1,∴A-2B=(2x2+3xy+2y-1)-2(x2-xy+x-12)=2x2+3xy+2y-1-2x2+2xy-

先化简再求值2(x2-xy)-3(2x2-3xy)-2[x2-(2x2-xy+y2)],其中x=-1,y=-2.

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.

如果x2+xy=2,xy+y2=-1,则x2-y2=______,x2+2xy+y2=______.

(1)∵x2-y2=x2+xy-xy-y2=x2+xy-(xy+y2)而x2+xy=2,xy+y2=-1,∴x2-y2=2-(-1)=3;(2)∵x2+2xy+y2=x2+xy+xy+y2,而x2+x

已知x2+4y2+x2y2-6xy+1=0,求 x4-y4/2x-y 乘 2xy-y2/xy-y2 除以(x2+y2/x

因为x²+4y²+x²y²-6xy+1=0(x²-4xy+4y²)+(x²y²-2xy+1)=0(x-2y)²