(2x-1)5=ax5 bx4 cx3 dx2 ex f 求b d值
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![(2x-1)5=ax5 bx4 cx3 dx2 ex f 求b d值](/uploads/image/f/11277-45-7.jpg?t=%282x-1%295%3Dax5+bx4+cx3+dx2+ex+f+%E6%B1%82b+d%E5%80%BC)
(x-6)(x^2+x+1)-x(x+1)(x-1)=x(2-5x)x^3+x^2+x-6x^2-6x-6-x^3+x=2x-5x^2-4x-6=2x6x=-6x=-1
是这样的:x^5+x^4=x^3(x^2+x)=(x^2+x)[(x^3-1)+1]=(x^2+x)(x^3-1)+x^2+x=[x(x+1)(x-1)](x^2+x+1)+x^2+x=(x^3-x)
这是两道题吗?1、两边同时乘(x-1)(x-2)(x-3)答案是x1=-1,x2=22、分母因式分解分别为(x-7)(2x+3)和(x-5)(x-7)然后同时乘(x-5)(x-7)(2x+3)答案是x
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
1)(x-3)/(x-2)-(x-5)/(x-4)=(x-7)/(x-6)-(x-9)/(x-8)化简得【(x-3)(x-4)-(x-5)/(x-2)】/【(x-2)(x-4)】=【(x-7)(x-8
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。。。
6/x+1=x-5/x^2+x6/x+1=x-5/x(x+1)两边同乘x(x+1)6x=x-5x=-1但是x(x+1)≠0x≠0或-1所以该方程无解
(X+2)/(X+1)-(X+4)/(X+3)=(X+6)/(X+5)-(X+8)/(X+7)(X+1+1)/(X+1)-(X+3+1)/(X+3)=(X+5+1)/(X+5)-(X+7+1)/(X+
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
我给个思路吧(X+2-1/X+2)-(X+3-1/X+3)=(X+6-1/X+6)-(X+7-1/X+7)1-1/(X+2)-1+1/(X+3)=1-1/(X+6)-1+1/(X+7)剩下的就自己解决
2x+4x+6x...+100x=1-(x+3x+5x+...+99x)x(2+4+6+...+100)=1-x(1+3+5+...99)x(2+4+6+...+100)+x(1+3+5+...+99
x/(x-2)=2x/(x-3)+(1-x)/(x-5x+6)x/(x-2)=2x/(x-3)+(1-x)/(x-2)(x-3)x(x-3)/(x-2)(x-3)=2x(x-2)/(x-2)(x-3)
/>(4X+2)/X+【4(X-5)-2】/(X-5)=【(X-4)-2】/(X-4)+【7(X+1)+2】/(X+1)4+2/X+4-2/(X-5)=1-2/(X-4)+7+2/(X+1)8+2/X
通分化简得到(x+1)(x+2)=(x+5)(x+6)3x+2=11x+308x=-28x=-7/2
首先由题意得x+1≠0,x+7≠0,x+5≠0,x+3≠0,即x≠-1,x≠-7,x≠-5,x≠-3,则先简化方程(x+1+1)/(x+1)+(x+7+1)/(x+7)=(x+5+1)/(x+5)+(
变形得9X—5=5X—1得4X=4得X=1再答:有不懂的再问再问:对啊……我就是想知道怎么变形……不过你这个比较详细一些……再问:右边是7-2变形,那左边是……再答:就是把未知数X移到一边再问:右边是
解(x-3)(x-7)/(x^2-8x+7)-(x-5)(x-1)/(x^2-8x+7)+(x^2-x)/(x^2-8x+7)=1x^2-10x+21-(x^2-6x+5)+x^2-x=x^2-8x+
X-1/X-2-(X-2)/(X-3)=(X-5)/(X-6)-(X-6)/(X-7)(X-1)(X-3)-(X-2)²/(X-2)(x-3)=(X-5)(X-7)-(X-6)²/
方程右边的1减到左边来,再同分得到(6x-2)/(x^2-x)=0得到x=1/3这个方程即使两边乘以x^2-x,x的最高次数为2次,也不会出现x^3,你肯定是什么地方弄错了