8(x 3)=10(x-3)
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(x3+5x2+4x-1)-(-x2-3x+2x3-3)+(8-7x-6x2+x3)=x3+5x2+4x-1+x2+3x-2x3+3+8-7x-6x2+x3=x3-2x3+x3+5x2+x2-6x2+
X³-3X=X(X²-3)=X(X+√3)(X-√3)
第二项应该是3x^2吧,先求导设k=3x^2+6x+6=3(x+1)^2+3当x=-1时斜率最小为3,切点为(-1,-14),所以切线方程为3x-y-11=0
(1)x3+x2+x-3=x^3-x+x^2+2x-3=x(x-1)(x+1)+(x-1)(x+3)=(x-1)(x^2+2x+3)(2)x3-6x2+11x-6=x^3-x^2-(5x^2-11x+
x3-3x2-11x+8=x(x2-3x)-11x+8=x-11x+8=-10x+8
(1)当a=2时,f(x)=2x+xlnx,f′(x)=−2x2+lnx+1,f(1)=2,f'(1)=-1,所以曲线y=f(x)在x=1处的切线方程为y=-x+3;(4分)(2)存在x1,x2∈[0
(2x3-3x2y-2xy2)-(x3-2xy2+y3)+(-x3+3x2y-y3)=2x3-3x2y-2xy2-x3+2xy2-y3-x3+3x2y-y3=-2y3=-2×(-1)3=2.因为化简的
f'(x)=3x^2-3x=1,f'(1)=0,f(1)=-2x=-1f'(1)=0,f(-1)=2x1,f'(x)>01>x>-1,f'(x)
这道题目分四步.第一步两边同加154得128x?÷8X3=2160第二步两边同除以3得128x?÷8=720第三步两边同乘以8128x?=5760第四步两边同除以128=45
原式=x3+5x2+4x-1+x2+3x-2x3+3+8-7x-6x2+x3=(x3-2x3+x3)+(5x2+x2-6x2)+(4x+3x-7x)+(-1+3+8)=0+0+0+10=10所以值恒不
2x3-3x+三分之二x3-2x-三分之四x3+x-1=4/3x3-4x-1=4/3x(-1/2)3+4x1/2-1=-4/3x1/8+2-1=1-1/6=5/6
原式=x3+5x2+4x-1+x2+3x-2x3+3+8-7x-6x2+x3=10,故与x无关.
原方程组可化为5x+3y=15 ①5x−3y=15 ②,①+②,得10x=30,x=3,代入①,得15+3y=15,y=0.则原方程组的解为x=3y=0.
f'(x)=3x²-3令f'(x)=0,解得x=1或x=-1令f('x)>0,解得x>1或x再答:不客气
根据题意:A=8x3-7x2+5x+(3x3-2x2-4)=8x3-7x2+5x+3x3-2x2-4=11x3-9x2+5x-4.
∵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
解法1:将常数项8拆成-1+9.原式=x3-9x-1+9=(x3-1)-9x+9=(x-1)(x2+x+1)-9(x-1)=(x-1)(x2+x-8).解法2:将一次项-9x拆成-x-8x.原式=x3
用比差法.(3x3-2x2-4x+1)-(3x3+4x+10)=-2x2-8x-9=-2(x2+4x)-9=-2[(x+2)2-4]-9=-2(x+2)2-1<0即(3x3-2x2-4x+1)-(3x
conv([1,8,0,0,-10],[2,-1,3])结果ans=215-524-2010-30即2*x6+15*x5-5*x4+24*x3-20*x2+10*x-30再问:如何输让215-524-