(6x^2+16x+18)/(x+1)(x+2)(x+3)
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/10/04 03:36:47
(6x^2+16x+18)/(x+1)(x+2)(x+3)
将下列各题分解为部分分式
将下列各题分解为部分分式
分解部分分式,一般采用待定系数法.
设(6x^2+16x+18)/(x+1)(x+2)(x+3)=a/(x+1)+b/(x+2)+c/(x+3)
=[(a+b+c)x^2+(5a+4b+3c)x+(6a+3b+2c)]/(x+1)(x+2)(x+3)
因此有:a+b+c=6,5a+4b+3c=16,6a+3b+2c=18
解这个三元一次方程组得:a=4,b=-10,c=12
所以(6x^2+16x+18)/(x+1)(x+2)(x+3)=4/(x+1)-10/(x+2)+12/(x+3)
设(6x^2+16x+18)/(x+1)(x+2)(x+3)=a/(x+1)+b/(x+2)+c/(x+3)
=[(a+b+c)x^2+(5a+4b+3c)x+(6a+3b+2c)]/(x+1)(x+2)(x+3)
因此有:a+b+c=6,5a+4b+3c=16,6a+3b+2c=18
解这个三元一次方程组得:a=4,b=-10,c=12
所以(6x^2+16x+18)/(x+1)(x+2)(x+3)=4/(x+1)-10/(x+2)+12/(x+3)
已知 X=4- 根号3 求 (X*X*X*X-6X*X*X-2X*X+18X+23) / (X*X-8X+15)的值
|X-1|+|X-2|+|X-3|+|X-4|+|X-5|+|X-6|+|X-7|+|X-8|+|X-9|+|X-10|
解方程x/(x-2)=2x/(x-3)+(1-x)/(x-5x+6)
(x+2/x^2-x-6)-(2x/x^2-3x)-(x+1/x^2-6x+9)
x+2/x^2-x-6 + 2x/x^3-3x - x+1/x^2-6x+9
化简:[√(x^2-6x+9)/x^2-x-12]*(x^3-16x)/(x^2-3x)-1/(x+3) {x>3}
x-1)(X-2)(x-3)...(x-50)+x(x-2)(X-3)...(X-50)+...+x(x-1)(x-2)
(x+1-1/1-x)÷(x-x²/x-1) (x-4/x-x-6+ x+2/x-3)÷x+1/x-3
1/x(x+3)+1/(x+3)(x+6)+1/(x+6)(x+9)=3/2x+18
x+2/x+1-x+3/x+2-x+4/x+3+x+5/x+4
x^5+x^4 = (x^3-x)(x^2+x+1)+x^2+x
约分 (X的平方+4x+3)(2x-x)^/(X^+X)(X^+X-6),