已知3X 2*5X 2=15x 2
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1/10x4+3x2+1=x4-x3+(x3+3x2+x)-x+1=x4-x3+x(x2+3x+1)-x+1=x4-x3-x+1=x4-(x3+3x2+x)+3x2+1=x4-x(x2+3x+1)+3
An=(2n-1)x2^n=nx2^(n+1)-2^n,则Sn=[nx2^2x(2^n-1)/(2-1)]-[2x(2^n-1)/2-1]=(2^n-1)(4n-1)
x2+3x2+1=0中的3x2表示什么?再问:已知X2+3X+1=0,求X2+1/X2的值?得数是7。求过程?我打的是X的平方。怎么会出X2、再答:答:因为x≠0,两边都除以x得:x+1/x=-3,两
∵x2+4x-5=0,∴x2+4x=5,∴3x2+12x-5=3(x2+4x)-5=3×4-5=7.故答案为7.
你可以参见“韦达定理”方程两个根的积是1,说明他们互为倒数.x^2+1/x^2=(x+1/x)^2-2*x*1/x=(-5)²-2=23
等式两边同时乘以(x+3)(x-2)(x+2)就可以去分母了
x^2+3x+1=0方程两边同除以xx+3+1/x=0x+1/x=-3x^2+1/x^2=(x+1/x)^2-2=(-3)^2-2=9-2=7
令t=x^2-2则x^2=t+2代入原式f(t)=lg((t+2)/(t-3))由于t和x只是字母而已,本质没有区别,所以解得f(x)=lg((x+2)/(x-3))^2表示平方然后就是((x+2)/
二十五分之二十四
x/x²-3x+1=1/5x²-3x+1=5xx²+1=8xx+1/x=8平方x²+2+1/x²=64x²+1/x²=62x
Sn=1*2+3*2^2+5*2^3+……+(2n-1)*2^n2Sn=1*2^2+3*2^3+...+(2n-1)*2^(n+1)相减得-Sn=1*2+2*2^2+2*2^3+..+2*2^n-(2
答:x²-5x=3(x-1)(2x-1)-x(x+3)+15/(x²-3)=2x²-3x+1-x²-3x+15/(5x)=x²-5x-x+1+3/x=
(x²+y²)²+(x²+y²)-6-6=0(x²+y²)²+(x²+y²)-12=0(x²
设x2=y3=z4=k(k≠0),则x=2k,y=3k,z=4k,所以,2x+3y−zx−3y+z=2•2k+3•3k−4k2k−3•3k+4k=9k−3k=-3.
x²-2xy+y²/x²-y²=(x-y)²/(x-y)(x+y)=(x-y)/(x+y)因为x=3,y=-5,所以(3-(-5))/(3+(-5))
22
已知X1X2为方程5X平方-3X-1=0两个根;所以x1+x2=3/5;x1x2=-1/5;x1-x2=√(x1-x2)²=√[(x1+x2)²-4x1x2]=√(9/25+4/5
x1³+x2³=(x1+x2)(x1²-x1*x2+x2²)=(x1+x2)[(x1+x2)²-3x1*x2]=3×(3²-3×1)=3×6
5x2-2x-15x2−2x−5=x+5-1x,∵5x2-3x=5,两边同除以5x得:x-1x=35,∴原式=x+5-1x=285.
X2-3X-1=0则X-3-1/X=0则X-1/X=3则(X-1/X)²=3²=9则X²-2+1/X²=9则X²+1/X²=9+2=11