# 方程式的解法-計算題兩題

(1)2/x三次方+1= a/x+1+ b(x+c)/x平方-x+1 則b=?

(2)設x+1/x(x-1)三次方 =a/x+ b/x-1 +c/(x-1)平方 +d/(x-1)三次方 且aˋbˋcˋd 均為常數 則a+2d=?

### 2 個解答

• 1 0 年前
最佳解答

(1)原式應該是2/(x三次方+1)= a/(x+1)+ b(x+c)/(x平方-x+1)吧

同乘(x三次方+1) => 2=a(x平方-x+1)+b(x+c)(x+1)

=> 2=(a+b)x平方+(b-a+bc)x+a+bc

所以 => a+b=0 , b-a+bc=0 , a+bc=2

=> a=2/3 , b=-2/3 , c=-2

(2)原式應該是(x+1)/x(x-1)三次方 =a/x+ b/(x-1)+c/(x-1)平方 +d/(x-1)三次方吧

同乘x(x-1)三次方 => x+1=a(x-1)三次方+bx(x-1)平方+cx(x-1)+d

=> x+1=(a+b)x三次方+(-3a-2b+c)x平方+(3a+b-c)x+(-a+d)

所以 => a+b=0 , -3a-2b+c=0 , 3a+b-c=1 , -a+d=1

=> a=1 , b=-1 , c=1 , d=2

=> a+2d=5

2006-07-27 22:45:07 補充：

抱歉 打錯(2)x+1=a(x-1)三次方+bx(x-1)平方+cx(x-1)+dx => x+1=(a+b)x三次方+(-3a-2b+c)x平方+(3a+b-c+d)x+(-a) => a+b=0 , -3a-2b+c=0 , 3a+b-c+d=1 , -a=1 => a=-1 , b=1 , c=-1 , d=2 => a+2d=3

參考資料： 自己
• 1 0 年前

2/(x^3+1)=a/(x+1)+b(x+c)/(x^2 -x+1)

=[a(x^2-x+1)+b(x+c)(x+1)]/(x^3+1)

=>ax^2-ax+a+bx^2+b(c+1)x+c=2

=>(a+b)x^2+(bc+b-a)x+bc+a=2

=>a+b=0

=>b=-a

=>bc+b-a=0 a+bc=2

=>a(1-c)=2

=>c=1-2/a

=>bc+b-a=-a(1-2/a)-2a=0

=>2=3a=>a=2/3

b=-2/3 c=-2

(2)

a/x+b/(x-1)+c/(x-1)^2+d/(x-1)^3

=[a(x-1)^3+bx(x-1)^2+cx(x-1)+dx]/x(x-1)^3

=[a(x^3-3x^2+3x+1)+bx^3-2bx^2+bx-cx^2-cx+dx]/x(x-1)^3

=[(a+b)x^3+(c-2b+3a)x^2+(b-c-3a+d)x+a]/x(x-1)^3

=x+1

=>a+b=0,c-2b+3a=0,b-c-3a+d=1,a=1

=>b=-1,c+5=0,d-c=5 a=1

=>a=1 b=-1 c=-5 d=0

2006-07-27 22:36:44 補充：

a+2d=2

2006-07-27 22:49:42 補充：

更正a/x+b/(x-1)+c/(x-1)^2+d/(x-1)^3=[a(x-1)^3+bx(x-1)^2+cx(x-1)+dx]/x(x-1)^3=[a(x^3-3x^2+3x+1)+bx^3-2bx^2+bx-cx^2-cx+dx]/x(x-1)^3=[(a+b)x^3+(c-2b-3a)x^2+(b-c+3a+d)x+a]/x(x-1)^3=x+1=>a+b=0,c-2b-3a=0,b-c+3a+d=1,a=1=>b=-1,c+1=0,d-c=-1 a=1=>a=1 b=-1 c=-1 d=0 =>a+2d=1