9^x-2·3^x 3k-1=0有两个实数解,求k的范围

x²+2x=3,x^2+2x-3=0,x1=1,x2=-3(舍）.（x²+6x+9）/（x²-9）·（x-3）/（x+3）-1/（x+3）=（x+3)^2/[(x-3)(

设a=(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)(x-7)(x-8)(x-9)那么y=a*(x-10);那么y^=a^*(x-10)+a*(x-10)^=a^*(x-10)+a那么y

 再问：额、不懂再答： 再答：后面的看做一个整体再问：好的吧、谢谢大神再答：回来的话，请采纳再问：啊、突然明白了呢。。。

(x+1)(x+2)(x+3)x+1=0(x^2+3x+2)(x^2+3x)+1=0[(x^2+3x)+2](x^2+3x)+1=0(x^2+3x)^2+2(x^2+3x)+1=0(x^2+3x+1)

x+x^2+x^3+x^4+x^5+x^6+x^7+x^8=（x+x^2+x^3+x^4）+（x^5+x^6+x^7+x^8）=x（1+x+x^2+x^3）+x^5（1+x+x^2+x^3）=（x+x

方程两边同时乘以x²-1:(3x²+9x+7)(x-1)-(2x²+4x-3)(x+1)-(x³+x+1)=03x^3+9x^2+7x-3x^2-9x-7-(2

x(2x-1)^2-4(x-3)(x^2+3x+9)-(2x+9)(9-2x)=0x(4x^2-4x+1)-(x^2+3x+9)(4x-12)-(81-4x^2)=04x^3-4x^2+x-(4x^3

1+x+x^2+x^3=0x+x^2+x^3+x^4+x^5+x^6+x^7+x^8=(x+x^2+x^3+x^4)+(x^5+x^6+x^7+x^8)=x(1+x+x^2+x^3)+x^5(1+x+

有种方法叫做穿针引线法,需要在纸上画出函数的简图,这个题的答案应该是-2,3,6,1

(x+2)/(x+3)-(x+1)/(x+2)=(x+8)/(x+9)-(x+7)/(x+8)(x+2)²/[(x+2)(x+3)]-(x+1)(x+3)/[(x+2)(x+3)]=(x+8

[(x^2)^2-2x^2*x+x^2]-(2x^2+2x+1)=0即(x^2-x)^2-(2x^2-2x)-4x+1=0[(x(x-1))^2-2(x(x-1))+1]-4x=0(x(x-1)-1)

同分化简得32x+128/(x+1)(x+3)(x+5)(x+7)=0所以得x=-4给分吧算了好久的····

x+3/x+1·2x的平方+6x/x的平方+6x+9-x/x+1=(x+3)/(x+1)·2x(x+3)/(x+3)²-x/(x+1)=2x/(x+1)-x/(x+1)=x/(x+1)=2/

（3m+1）（3n+2）=9mn+3m+3n+2=9mn+3m+3n+3-1=3(3mn+m+n+1)-1所以选B

1/x+1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+...1/(x+9)(x+10)=01/x+1/x-1/(x+1)+1/(x+1)-1/(x+2)+1/(x+2)-1/(

K+1=1

1,应该是x是几次幂就有几个根吧,所以应该是7+2=9个2.-7.036622145,-6.489288572,-6.200988153,-4.734373320,-4.289168546,-2.77

每两项合并,即(1+x)+(1+x)*x^2+(1+x)*x^4+(1+x)*x^6+(1+x)*x^8=0提取公因式(1+x),得到(1+x)(1+x^2+x^4+x^6+x^8)=0等式左边的两个

2x/x^2-9=x-1/x^2-6x+9-1/x+32x/(x+3)(x-3)=(x-1)/(x-3)²-1/(x+3)两边同乘以(x-3)²(x+3)2x(x-3)=(x-1)

=x^2002(x^4+x^3+x^2+x+1)=0提取公因式就行了