方程[x加1]][x加2][x加3][x加4]减8等于0-搜答案-智慧作业帮

令a=(x²+1)/x则x/(x²+1)=1/a所以a+2/a=3a²+2=3aa²-3a+2=0(a-1)(a-2)=0a=1,a=2(x²+1)/

x２+2+1/x=2(x+1/x)x２+2+1/x=2x+2/xx２-2x+2=1/x画图象,得x=1

因式分解为（3x+2）(x-4)=0故x=-2/3x=4

再问：加3呢再答：那儿加3再答：那个3，不是分母吗？再答：采纳啊

(5x+2)/(x^2+x)=3/(x+1)(5x+2)/x=35x+2=3x2x+2=0x=-1不知原式写对了没有?

3x+2=4x+13x-4x=1-2-x=-1x=1

(x^2-3x+1)/(x^2-4)=30/(x+2)(x^2-3x+1)/(x-2)=30x^2-3x+1=30x-60x^2-33x+61=0x=(33±√1333)/2

(x^2-3x+1)/(x^2-4)=30/(x+2)(x^2-3x+1)/(x-2)=30x^2-3x+1=30x-60x^2-33x+61=0x=(33±√1333)/2

3x+2x/3=14;11x/3=14;x=42/11;0.6x+5=65;0.6x=60;x=100;x/2+x/3=3;5x/6=3;x=18/5;如果本题有什么不明白可以追问,如果满意记得采纳如

原方程为:x+x\2+1-x+x\4+3=x+x\6+5-x+x\8+7整理得：3x\4+4=7x\24+12两边同时乘以24,得18x+96=3x+288,x=12.8

(x+1)(x+2)(x+3)(x+4)+1=[(x+1)(x+4)][(x+2)(x+3)]+1=(x^2+5x+4)(x^2+5x+6)+1把x^2+5x看做A,所以原式化成：(A+4)(A+6)

2x-(x+1)÷2＝2+（x+1）÷22x-0.5x-0.5=2+0.5x+0.51.5x-0.5=0.5x+2.51.5x-0.5x=2.5+0.5x=3

X+(X+2)+X/4=742X+X/4=728X+x=2889X=288X=32

(x-1)(x+2)(x+3)(x+6)+20=0(x-1)(x+6)(x+2)(x+3)+20=0(x^2+5x-6)(x^2+5x+6)+20=0(x^2+5x)^2-36+20=0(x^2+5x

8x+22=2(6x+1)8x+22=12x+24x=20x=5

2x平方?如果是那解就是x①=-0.027820777536819108x②=-17.972179222463182

x(x+1)+(x+1)/x^2=0(x+1)(1/x^2+x)=0x=-1

(x+1)(x+2)+1/4=x²+3x+2+1/4=x²+3x+9/4=x²+2*3x/2+(3/2)²=(x+3/2)²

1+1/(x+2)+1+1/(x+3)=1+1/(x+4)+1+1/(x+1)1/(x+2)+1/(x+3)=1/(x+4)+1/(x+1)1/(x+3)-1/(x+4)=1/(x+1)-1/(x+2

x=1再问：给个过程吧