已知A=3x3一2mx 3x 1,
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x=log2(y)则X1+2X2+3X3=log2(y1)+2log2(y2)+3log2(y3)=log2(y1)+log2(y2^2)+log2(y3^3)=log2(y1y2^2y3^3)=1所
A-(B+C)=(x3-2x2+4x+3)-[(x2+2x-6)+(x3+2x-3)]=(x3-2x2+4x+3)-[x2+2x-6+x3+2x-3]=x3-2x2+4x+3-x2-2x+6-x3-2
1.对函数求一阶导:令y=f(x)'=3x(x-a),得到极值点x=0或x=a2.由于a>1;则f(x)在x=0取最大值1,在-1或1处取最小值-2,(题上区间应该是【-1,1】吧?3.由2则f(0)
因为A+B+C=x3-2y3+3x2y+xy2-3xy+4+y3-x3-4x2y-3xy-3xy2+3+y3+x2y+2xy2+6xy-6=1,所以,对于x、y、z的任何值A+B+C是常数.
(Ⅰ)由题设可知:f'(1)=0且f(1)=2,即3−6a−b=01−3a−b=2,解得a=43,b=−5.;(Ⅱ)∵f'(x)=3x2-6ax-b=3x2-6ax-9a,又f(x)在[-1,2]上为
表达式有问题.x2+3x2+1=0,哪个是方哪个是系数?再问:右边的是方,左边的是系数再答:x2+3x2+1=0,4x2=-1? 式子没有意义啊再问:额,我不是这个意思那个是x不是乘号,3x2,这个是
f(x)=x^3+(1-a)x^2-a(a+2)x+b^表示次方1)函数f(x)的图象过原点,那么f(0)=0所以0=0+bb=0f'(x)=3x^2+2(1-a)x-a(a+2)f'(0)=-a(a
|X1-1|+|X2-2|+|X3-3|+...+|X2005-2005|=0所以|X1-1|=0,|X2-2|=0,|X3-3|=0,...,|X2005-2005|=0即x1=1,x2=2,x3=
A+B+C=(x3+3x2y-5xy2+6y3-1)+(y3+2xy2+x2y-2x3+2)+(x3-4x2y+3xy2-7y3+1)=(1+1-2)x3+(3+1-4)x2y+(-5+2+3)xy2
第一步先求导f^(X)=-3x2+6x+9第二步令导数f^(x)=-3x2+6x+9=0得x1=3,x2=-1对于导数f^(x)当f^(x)>0时可得x的范围为{-1
∵A=1+2x2-3x3,B=3x3-2x2-5x-4,∴2A-(A-B)=A+B=(1+2x2-3x3)+(3x3-2x2-5x-4)=1+2x2-3x3+3x3-2x2-5x-4=-3-5x.当x
(Ⅰ)由题意得f′(x)=3x2-2ax-3,∵f(x)在区间[1,+∞)上是增函数,∴当x∈[1,+∞)时,恒有f′(x)≥0,即3x2-2ax-3≥0在区间[1,+∞)上恒成立,由△=4a2+36
1.=2y1-5y'2>=3y1+y'2>=-5y1无限制,y2>=02.
1a=1/4f(x)=-2/3x³+1/2x²+3xf'(x)=-2x²+x+3令f'(x)=0即2x²-x-3=0解得x1=-1,x2=3/2随x在[-2,2
(1)f'(x)=2x2-4ax-3,对称轴x=a∈[−14,14]⊂(−1,1)f′(x)min=f′(a)=−2a2−3<0,f′(1)=−2a−72<0,f′(−1)=2a−72<0f′(x)m
根据题意:A=8x3-7x2+5x+(3x3-2x2-4)=8x3-7x2+5x+3x3-2x2-4=11x3-9x2+5x-4.
(Ⅰ)f′(x)=x2-4ax+3a2=(x-a)(x-3a),因为a>1,所以3a>a,∴f(x)的极小值为f(3a)=-1(Ⅱ)若1<a≤2时,当x∈[-1,a]时f/(x)>0,f(x)在[-1
∵A=x+2x2-3x3,B=3x3-3x2-x-4,∴2A-(A-B)=2A-A+B=A+B=x+2x2-3x3+3x3-3x2-x-4=-x2-4,∴当x=-23时,原式=-(-23)2-4=-4
A-B=(x3+2y3-xy2)-(﹣y3+x3+2xy2)=x³+2y³-xy²+y³-x³-2xy²=3y³-3xy²