-3x^(m 4)y^2-m
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原式=(y-x)2m[(x-y)m+(y-x)m]讨论:当m为偶数时,原式=2(y-x)3m;当m为奇数时,原式=0.
[x-(m+3)]^2+[y+(1-4m^2)]^2=-16m^4-9+(m+3)^2+(1-4m^2)^2所以r^2=-16m^4-9+(x+3)^2+(1-4m^2)^2>0-16m^4-9+(m
(1)当y=0时,14x2+(m4+1)x+m=0,(1分)x2+(m+4)x+4m=0,x1=-4,x2=-m.(2分)∵m<4,∴A(-4,0),B(-m,0)(5分)(2)过点C作CD⊥x轴,垂
圆的方程化为[x-(m+3)]2+[y-(1-4m2)]2=1+6m-7m2,则有1+6m-7m2>0,解得-17<m<1,故m的取值范围是(-17,1).
根据圆的方程,得出楼上正解.
(1)由方程x2+y2-2(m+3)x+2(1-4m2)y+16m4+9=0变形得:[x-(m+3)]2+[y+(1-4m2)]2=-7m2+6m+1,当且仅当-7m2+6m+1>0,即7m2-6m-
已知x^(3m)=2y^(2m)=3(x^(2m))^3+(y^m)^6-(x^2*y)^3m*y^m=x^6m+y^6m-x^6my^4m=(x^3m)^2+(y^2m)^3-(x^3m)^2*(y
y=x-3x+1经过点(m,0),代入得到m^2-3m+1=0m^2=3m-1m4-21m+10=(3m-1)^2-21m+10=9m^2-6m+1-21m+10=9m^2-27m+11=9(3m-1
已知m是方程x²+2x-5=0的一个根则有m²+2m-5=0,m²+2m=5m4+4m3+4m²+1=(m²+2m)²+1=25+1=26
方程两边都乘(x-4),得2=3(x-4)-m∵当最简公分母x-4=0时,方程有增根,∴把x-4=0代入整式方程,∴m=-2.故选A.
[x-(m+3)]^2+[y-(1-4m^2)]^2=-(7m^2-6m-1)=-(7m+1)(m-1)>0-1/7再问:半径是多少再答:根号下[-(7m+1)(m-1)]
(1)由方程x2+y2-2(m+3)x+2(1-4m2)y+16m4+9=0变形得:[x-(m+3)]2+[y+(1-4m2)]2=-7m2+6m+1,要使方程表示圆,则需要-7m2+6m+1>0;如
去分母得,5x-3m=2m-15,移项得,5x=2m+3m-15,合并同类项得,5x=5m-15,系数化为1得,x=m-3,∵x是正数,∵m-3>0,解得m>3.
解方程4(3x-7)=19-35x得:x=1,将x=1代入3m+5x6=m4−74x得:3m+56=m4-74,解得:m=-313.
[(y-x)^3]*[(x-y)^m]-[(x-y)^(m+2)]*(y-x)=-[(x-y)^3]*[(x-y)^m]+[(x-y)^(m+2)]*(x-y)=-[(x-y)^(m+3)]+[(x-
M是方程x2-x-1=0的是一个根,所以:M^2-M-1=0M^2=M+1M^4-2M^3+M^2=(M+1)^2-2M(M+1)+M+1=M^2+2M+1-2M^2-2M+M+1=-M^2+M+1+
3x+2y=m+1①4x+y=m②①-②×2得:-5x=-m+1,解得:x=m−15,①×4-②×3得:5y=m+1,解得:y=m+15,即方程组的解是x=m−15y=m+15,∵关于x、y的方程组3
(1)x2+y2-2(m+3)x+2(1-4m2)y+16m4+9=0x2-2(m+3)x+(m+3)2+y2+2(1-4m2)y+(1+4m2)2=(m+3)2+(1+4m2)2-16m4-9(x-