6x^2-5xy y^2-8x 2y-8=0的图象与X轴围成的图形的面积-搜答案-大师作文网

原式=[x-y(x-y)2-y(x+y)(x+y)(x-y)]•xyy-1=（1x-y-yx-y）•xyy-1=1-yx-y•xyy-1=-xyx-y．故答案是：-xyx-y．

设(x²-1)/(x²+2x)=t则8t+3/t=118t²-11t+3=0(8t-3)(t-1)=0解得t=3/8或t=11.t=3/8(x²-1)/(x&#

后面的x²＋11x－708有误吧!再问：没有题目就这样能不能帮我再答：那我就试试：原式为：1/x2+x+1/x2+3x+2+1/x2+5x+6+1/x2+7x+12+1/x2+9x+20=5

2X-2.8x2.5=8.962x=8.96+7=15.96x=7.988X+2X=31.410x=31.4x=3.142x(X+1)=6x^2+x-3=0x=(-1+√13)/2或(-1-√13)/

原式=(x²+3x+9)/(x-3)(x²+3x+9)-6x/x(x-3)(x+3)-(x-1)/2(x+3)=1/(x-3)-6/(x-3)(x+3)-(x-1)/2(x+3)=

原式=[1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)]×(x+4)(x+5)=[1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+1/(x+3)-1/(x

分解因式下：X4-2X2-8=(x²-4)(x²+2)=(x+2)(x-2)(x²+2)X2-3XY+2Y2=(x-2y)(x-y)2X2-7X+3=(2x-1)(x-3

1/(x²+3x+2)=[(x+2)-(x+1)]/(x+1)(x+2)=1/(x+1)-1/(x+2)同理1/(x²+5x+6)=1/(x+2)-1/(x+3)1/(x²

1/(x2-5x+6)-1/(4x-x2-3)-1/(3x-x2-2)=1/(x2-5x+6)+1/(x2-4x+3)+1/(x2-3x+2)=1/(x-2)(x-3)+1/(x-3)(x-1)+1/

等式两边同时乘以(x+3)(x-2)(x+2)就可以去分母了

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原式=[(x+y)2(x-y)(x+y)+-4xy(x-y)(x+y)]×(x+3y)(x-3y)(x+3y)(x-y)=x-3yx+y，由已知得（3x-2y）（x+y）=0，因为x+y≠0，所以3x

产生4种配子xyxyyyx有一种y有两种xy1y2xy1xy2y1y2产生的配子如上y1、y2是一个意思所以分为一组xyxyyy=1221

1/(x2+x)+1/(x2+3x+2)+1/(x2+5x+6)+1/(x2+7x+12)=1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)=1/x-1/

原式=（5-3-2）x2+（-5+6）x+（4-5）=x-1

y=2x-5所以x+2y=x+4x-10=5a5x=5a+10x=a+2y=2x-5=2a-1xyy所以x>0>y所以a+2>00>2a-1,a-1/2

原式=x3+5x2+4x-1+x2+3x-2x3+3+8-7x-6x2+x3=10，故与x无关．

原式=5x²-x²-(4x-x²)+2(x²-3x)=4x²-4x+x²+2x²-6x=7x²-10x

x2-5x+8=0a=1,b=-5,c=8∴△=b²-4ac=(-5)²-4×1×8=25-32=-7

可以解得B={2,3},C={-4,2}根据条件A∩B≠空集,且A∩C=空集得3为A中的一个解2不是A的解代入得：9-3a+a2-19=0a1=5a2=-2而4-2a+a2-19不等于0a不等于5或-