已知[1-2x]的五次幂
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x^2-3x+1=0两边除xx-3+1/x=0x+1/x=3平方x^2+2+1/x^2=9x^2+1/x^2=7平方x^4+2+1/x^4=49x^4+1/x^4=47(x+1/x)(x^2+1/x^
x=1时,右边=a+b+c+d+e+f所以a+b+c+d+e+f就是(2x-1)5次方当x=1时的值所以a+b+c+d+e+f=(2×1-1)5次方=1
1.-32.第二题题目不太清楚,是y^2(n-1)还是y^(2n-1),如果是前者,n=2,如果是后者,n=3/2.3.x=1/24.-25.46.1200平方米7.盈利6元(还有,应该是都卖64元吧
x³+y³=(x+y)(x²-xy+y²)=x²-xy+y²=(x+y)²-3xy=1-3xy=1/3∴xy=2/9x五次方+y五
以(2x-1)^5表示2x-1的五次方,依此类推(2x-1)^5=ax^5+bx^4+cx^3+dx^2+ex+f当x=1时,1=a+b+c+d+e+f...①当x=-1时,(-3)^5=-a+b-c
x的四次方-4x的三次方+4x²+251=(2x+1)²-4x(2x+1)+4x²+251=4x²+4x+1-8x²-4x+4x²+251=
x^5--(m--2)x^4+5x^3+(n--1)x^2---1不含x的四次方和x的平方项,m==2;n==1
x²=x+1两边平方x^4=x²+2x+1=(x+1)+2x+1=3x+2x^5=x×x^4=x(3x+2)=3x²+2x=3(x+1)+2x=5x+3所以原式=(5x+
x²=x+1平方x^4=x²+2x+1=(x+1)+2x+1=3x+2x^5=x^4*x=x(3x+2)=3x²+2x=3(x+1)+2x=5x+3所以原式=(3x+2+
x²-x-1=0x²=x+1x^4=(x+1)²=x²+2x+1=(x+1)+2x+1=3x+2x^5=x^4*x=(3x+2)*x=3x²+2x=3
∵x²-3=0∴x²=3,x=±√3∴(2x-1)²+(x+2)(x-2)-(x^5-4x^4)+x³=4x²-4x+1+x²-4-x(x&
证明:x+1=(x+1)(x^2-x+1)=(x+1)(1+1)=2x+2所以x=2x+1x^5+x^2=x^2(x+1)=x^2(2x+2)=2x^3+2x^2=2(2x+1)+2x^2x^5=4x
(x^2+x+!)(x^3-x^2+1)
X=1.13299756588507
根据题意x²=3x=±√3所以代数式=4x²-4x+1+x²-4-x²*x^3+(2x²)²+x^3=12-4x+1+3-4-3x^3+36
x^2=x+1x^5-5x+2=x(x^2)^2-5x+2=x(x+1)^2-5x+2=x^3+2x^2+x-5x+2=x(x+1)+2x^2-4x+2=3x^2-3x+2=3x+3-3x+2=5
(a-2)x^2y|a|+1的次数为2+|a|+1=5,a=±2又系数a-2≠0,则a=-2