若a4+b4=2a2-2a2b2+2b2-1
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∵|a+b+5|+(a+2)2=0,∴a+b+5=0,a+2=0,解得:a=-2,b=-3,∴3a2b-[2a2b-(3ab-a2b)-4a2]-2ab=3a2b-[2a2b-3ab+a2b-4a2]
(b1,b2,b3,b4)=(a1,a2,a3,a4)A矩阵A=1001110001100011这里有个结论:r(b1,b2,b3,b4)=r(A)下面计算A的秩r1-r2+r3-r400001100
∵c4-2(a2+b2)c2+a4+a2b2+b4=0,⇒c4-2(a2+b2)c2+(a2+b2)2-a2b2=0,⇒[c2-(a2+b2)]2-(ab)2=0,⇒(c2-a2-b2-ab)(c2-
∵(a+2)2+|a+b+5|=0,∴a+2=0a+b+5=0,解得a=−2b=−3,∵原式=3a2b-2a2b+2ab-a2b+4a2-ab=(3-2-1)a2b+ab+4a2=4a2+ab=a(4
4=b1-b2+b3所以线性相关
(a-b)c3-(a2-b2)c2-(a3-a2b+ab2-b3)c+a4-b4=(a-b)c3-(a-b)(a+b)c2-(a2*(a-b)+b2*(a-b))c+(a-b)(a+b)(a2+b2)
[b1,b2,b3,b4]=[1100,0110,0011,1001][a1,a2,a3,a4]求[1100,0110,0011,1001]的行列式,如果等于0,那么线性相关如果不等于0,那么线性无关
4=b1+b3-b2故b1,b2,b3,b4线性相关.
a^4+b^4=(a²+b²)²-2a²b²+ab=1-2(ab)²+ab设x=ab,则有f(x)=-2x²+x+1很显然,该函数
首先我们先算出总和即1+2+3+4+5+6+7+8=29,说明A的最大和为14,设a1
(1)设{an}的公比为q,∵a1=2,a4=54,∴q=3,∴an=2•3n−1,Sn=2(1−3n)1−3=3n−1; (2)设{bn}的公差为d,则4b1+6d=27-1=
∵∴∵a+b+c=2∴4=(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ac=3+2(ab+bc+ac)∴ab+bc+ac=1/2(1)∵a+b+c=2∴8=(a+b+c)^3=a^3
a4+b4+c4=2c2(a2+b2),a4+b4+c4-2c2(a2+b2)=0a^4+b^4+c^2-2a^2c^2-2b^2c^2+2a^2b^2=2a^2b^2(c^2-a^2-b^2)^2=
c^4-2c^2(a^2+b^2)+(a^2+b^2)^2-a^2b^2=0(c^2-a^2-b^2)^2=a^2b^2a^2+b^2-c^2=ab或-abcosC=(a^2+b^2-c^2)/(2a
(a2+b2)(a2+b2-2)=15,把a2+b2看一整体,可解得a2+b2=5或-3(平方和不能为负,-3舍掉)最后的结果是5
∵c4-2(a2+b2)c2+a4+a2b2+b4=0,⇒c4-2(a2+b2)c2+(a2+b2)2-a2b2=0,⇒[c2-(a2+b2)]2-(ab)2=0,⇒(c2-a2-b2-ab)(c2-
(a^4+b^4)(a^2+b^2)-(a^3+b^3)^2=a^6+a^2b^4+a^4b^2+b^6-a^6-2a^3b^3-b^6=a^2b^4+a^4b^2-2a^3b^3=(a^2b^2)(
a^4+b^4+c^4-2c^2(a^2+b^2)=0(a^2+b^2-c^2)2=a^4+b^4+c^4-2c^2(a^2+b^2)+2a^2b^2(a^2+b^2-c^2)2=2a^2b^2cos