已知A=3x2-xy y2,B=2x2
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解题思路:该试题考查集合的基本运算,以及二次方程的求解解题过程:
这俩式子相加得A-C=-x-1所以C-A=x+1
x1+y1=-2x2+y2=-3再问:有详细点步骤么
∵A=x2+xy+y2,B=-3xy-x2∴2A-3B=2(x2+xy+y2)-3(-3xy-x2)=2x2+2xy+2y2+9xy+3x2=5x2+11xy+2y2.
由A中不等式变形得:(x-1)(x+6)≤0,解得:-6≤x≤1,即A=[-6,1];由B中不等式变形得:x(x+3)≥0,解得:x≤-3或x≥0,即B=(-∞,-3]∪[0,+∞),则A∩B=[-6
a*b=-6|a||b|=6得ab夹角是180度设a=kb(k
A=2x-1/x2+3x+2>0,B=x2+ax+b小于等于0是(2x-1)/(x^2)+3x+2>0x^2+ax+b小于等于0么?再问:A=2x-1/x2+3x+2>0;B=x2+ax+b小于等于0
因为|A|=2,|B|=3所以设x1=2coaa,Y1=2sina,X2=3cosb,Y2=3sinb因为A*B=6cosacosb+6sinasinb=6cos(a-b)=-6所以cos(a-b)=
A•(C-B)=(3x2-2x-1)[(x2-3x+6)-(x2-4x+7)]=(3x2-2x-1)[x2-3x+6-x2+4x-7]=(3x2-2x-1)(x-1)]=3x3-2x2-x-3x2+2
A:x²-3x+2=0解得x=1或x=2B:△=a²-4(a-1)=(a-2)²≥0,所以B不为空集再问:当b属于A时,求实数a的取值集合?再答:哪儿来的b,你问什么?再
A+B+A-B=4+3X-4-X+3,2A=3+2X,A=2/3+XA+B-A+B=4+3x-4+X-3,2B=4X-3,B=2X-2/3再问:已知A+B=2x2+3x-4,A-B=-x2+3,求A,
∵x2-16<0⇒-4<x<4,∴A={x|-4<x<4},∵x2-4x+3>0⇒x>3或x<1,∴B={x|x>3或x<1},∴A∩B={x|-4<x<1或3<x<4}A∪B=R
(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
∵f(x)=x^2+4x+3∴f(ax+b)=(ax+b)^2+4(ax+b)+3=a^2x^2+(4a+2ab)x+b^2+4b+3=x^2+10x+24两个多项式相等,那么对应系数相等∴a^2=1
(1)A=4(2-x2)-2x,B=2x2-x+3.A-2B=4(2-x2)-2x-2(2x2-x+3)=-8x2+2当x=14时,A-2B=-8×(14)2+2=32;(2)A=4(2-x2)-2x
(1)2A-B=2(3x2+3y2-5xy)-(2xy-3y2+4x2)=6x2+6y2-10xy-2xy+3y2-4x2=2x2+9y2-12xy;(2)当x=3,y=−13时,2A-B=2x2+9
f(x)=x^3+1.5(1-a)x^2-3ax+b吧.1.f'(x)=3x^2+3(1-a)x-3a=3(x+1)(x-a),由a>0可知f'(x)>0的解为x>a或者x
(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-
(3A-2B)-(2A+B)=3A-2B-2A-B=A-3B,将A、B代入,即得:4x2-4xy+y2-3(x2+xy-5y2)=4x2-4xy+y2-3x2-3xy+15y2=x2-7xy+16y2
对于集合A:x2-9≤0,化为(x-3)(x+3)≤0,解得-3≤x≤3,∴集合A=[-3,3];对于集合B:x2-4x+3>0,化为(x-3)(x-1)>0,解得3<x或x<1,集合B=(-∞,1)