x2−m2 4mn−4n2
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设f(x)=x^2-(m^2+n^2-6n)x+m^2+n^2+2m-4n+1两个实数根满足x1
mx3-2x2+3x-4x3+5x2-nx=(m-4)x3+3x2+(3-n)x,因为不含三次项及一次项的多项式,依题意有(1)m-4=0,m=4;(2)3-n=0,n=3.代入m2-mn+n2,原式
∵x2-4x+y2+6y+z−3+13=0,∴(x-2)2+(y+3)2+z−3=0,∴x-2=0,y+3=0,z-3=0,解得x=2,y=-3,z=3,∴(xy)z=[2×(-3)]3=-216.
(1)原式=x(x+9)x(x+3)+(x+3)(x−3)(x+3)2=x+9x+3+x−3x+3=2(x+3)x+3=2;(2)原式=-x−2x−1÷x2−4x−1=-x−2x−1•x−1(x+2)
∵−12≤x≤1,∴x-1≤0,x-3<0,2x+1≥0,∴x2−2x+1+x2−6x+9+4x2+4x+1=(x−1)2+(x−3)2+(2x+1)2=|x-1|+|x-3|+|2x+1|=1
m²(n²-1)+4mn-n²+1=m²n²-m²+4mn-n²+1=(m²n²+2mn+1)-(m²
原式=((m-n)x-(m+n))((m+n)x+(m-n))=0x=(m+n)/(m-n)或-(m-n)/(m+n)再问:能说清楚点吗?再答:(m+n)(m-n)x2-4nmx-(m+n)(m-n)
方程的根是:x=[-b±√(b²-4ac)]/2a∴x1+x2=[-b+√(b²-4ac)]/2a+[-b-√(b²-4ac)]/2a=-2b/2a=-b/a方程两个根是
m,n是方程x2+3x+1=0的两根由韦达定理得:m+n=-3且m²+3m+1=0m²=-3m-1同理:n²=-3n-1则:2m2+4n2-6n+2000=2(3m-1)
∵x2-9=0,∴x=±3,当x=3时,x2-4x+3=0,∴x=3不满足条件.当x=-3时,x2-4x+3≠0,∴当x=-3时分式的值是0.故选C.
(m²-n²)x²-4mnx-(m²-n²)=0(m+n)(m-n)x²-4mnx-(m+n)(m-n)=0[(m-n)x-(m+n)][(
设x2-3x=y,则原方程可化为:y+3y=4.即:y+3y−4=0.故选A.
x=2n/1+n2,y=1-n2/1+n2x2+y2=[(4n^2+(1-n^2)^2]/(1+n^2)^2=(1+n^2)^2/(1+n^2)^2=1【欢迎追问,】
limn→∞(1+2+…+nn+2−n2)=limn→∞ (n(1+n)2n+2−n2)=limn→∞−n2(n+2)=−12故答案为:−12
由x1≤0及0≤x2≤1∴x1+x2=m²+n²-6n≤1(1)x1×x2=m²+n²+2m-4n+1≤0(2)由(2)(m²+2m+1)+n&sup
三角代换,令x=根号2*cosa,y=根号2*sina;m=2*cosb,y=2*sinb;则xm+yn=2倍根号2*(cosacosb+sinasinb)=2倍根号2*cos(a-b).故最大值就是
整理得:2m−5n=20①2m+3n=4②,①-②得:-8n=16,∴n=-2,把n=-2代入②得:2m-6=4,∴m=5,∴方程组的解是m=5n=−2.
由题意可得|m+4|+(n-1)2=0,∴m+4=0n−1=0,解得m=−4n=1,∴x2+4y2-mxy-n,=x2+4y2+4xy-1,=(x+2y)2-1,=(x+2y+1)(x+2y-1).
问题没说完呢..再问:mx+ny的最大值。再答:做变量替换:x=(3)^(1/2)*sin(u)y=(3)^(1/2)*cos(u)m=sin(v)n=cos(v)则mx+ny=3^(1/2)*(si
x²-2mx=n²-m²(x-m)²-m²=n²-m²(x-m)²=n²x-m=±nx=m+n或x=m-n如还