已知4x 3=1时二那么4x-3等于多少
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原式=3x3-(x3+6x2-7x)-2x3+6x2+8x,=3x3-x3-6x2+7x-2x3+6x2+8x,=15x,当x=-1时,原式=15x=15×(-1)=-15.
x³-2x²-4x-5=(x³-2x²+x)-(5x-5)-10=x(x-1)²-5(x-1)-10将x=1+√5代入原式=(1+√5)(1+√5-1
已知,X2+X-1=0----1)X2=1-X-----------2)X3-2X+4=X(X2-2)+4将2式代入,=X(1-X-2)+4=-X2-X+4再将2式代入,=X-1-X+4=3
x4+2x3+4x2+3x+2=(x4+x3+x2)+(x3+x2+x)+(2x2+2x+2)=x²(x²+x+1)+x(x²+x+1)+2(x²+x+1)=(
(x1+x2+x3+x4+x5)/5=20x1+x2+x3+x4+x5=100(x1+x2+1+x3+2+X4+3+X5+4)/5=(x1+x2+x3+x4+x5+10)/5=110/5=22
因为X^5+X^6+X^7+X^8+X^9=X^5(1+X^1+X^2+X^3+X^4)=0..X^1995+X^1996+X^1997+X^1998+X^1999=X^1995(1+X^1+X^2+
f(x)=x²-x-5g(x)=1/3x³-5/2x²+4xg'(x)=x²-5x+4y=g'(x)/[f(x)+9]=(x²-5x+4)/(x
∵x2+x-1=0,∴x2=1-x,x2+x=1,∵x3-2x+4,=x(x2-2)+4=x(1-x-2)+4=x(-1-x)+4=-x2-x+4,=-(x2+x)+4=3.故答案为:3.
X2-2x-1=0(x-1)2-2=0(x-1)2=2x-1=±√2x=1±√2代入2x3-3x2-4x+2解得即可
(Ⅰ)f′(x)=x2+2ax+4,∴f′(0)=4,且f(0)=b;∴在点(0,f(0))处的切线方程为:y=4x+b;解y=4x+by=13x3+ax2+4x+b得:x=0,或x=-3a;∵a≠0
53/582再问:怎么算的啊?再答:x7=85/6x1+x2.....+x10=971/685/6/971/6=53/582
∵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
∵3x3-x=1,∴9x4+12x3-3x2-7x+2001,=3x(3x3-x-1)+4(3x3-x-1)+2005,=2005.故选D.
y’=(4x^3-5x^2+3x-2)'=12x^2-10x+3y"=(12x^2-10x+3)'=24x-10y"(0)=24*0-10=-10
(1)对f(x)求导得:f(x)'=3X^2-8X+4令f(x)>0得:x>2或x
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
∵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
解题思路:函数性质一定要好好使用。围绕单调性、奇偶性、周期性以及特殊点做文章。解题过程:答案见附件,有问题请在讨论区交流。最终答案:略
根据题意列得:(ax2+bx+1)(3x+1)=3ax3+(a+3b)x2+(b+3)x+1,∵不含x3的项,也不含x的项,∴3a=0,b+3=0,则a=0,b=-3.故答案为:0;-3
8*7-1-2-3-4=4642a