a/(2x-1) - b/(x+3) + 2/(2x^2+5x-3) = 9/(2x-1)(x+3)
find constants a and b
(a=2,b=1)
please show the steps
數學 三元一次方程式
2010-01-14 5:15 am
回答 (4)
2010-01-14 5:22 am
✔ 最佳答案
a/(2x-1) - b/(x+3) + 2/(2x^2+5x-3) = 9/(2x-1)(x+3)a/(2x-1) - b/(x+3) + 2/(2x-1)(x+3) = 9/(2x-1)(x+3)
Multiply both sides by (2x-1)(x+3),
(x+3)a - (2x-1)b + 2 = 9
Put x = -3,7b+2=9
b=1
Put x = 1/2,7a/2 + 2 = 9
a = 2
2010-01-14 5:56 am
a/(2x-1) - b/(x+3) + 2/(2x^2+5x-3) = 9/(2x-1)(x+3)
a/(2x-1) - b/(x+3) + 2/(2x-1)(x+3) = 9/(2x-1)(x+3)
multiply (2x-1) on both sides and let x=1/2,a=7/(7/2)=2
multiply (x+3) on both sides and let x=-3,b=1
a/(2x-1) - b/(x+3) + 2/(2x-1)(x+3) = 9/(2x-1)(x+3)
multiply (2x-1) on both sides and let x=1/2,a=7/(7/2)=2
multiply (x+3) on both sides and let x=-3,b=1
2010-01-14 5:35 am
a/(2x-1) - b/(x+3) + 2/(2x-1)(x+3) = 9/(2x-1)(x+3)
multiply (2x-1) on both sides and let x=1/2,a=7/(7/2)=2
multiply (x+3) on both sides and let x=-3,b=1
multiply (2x-1) on both sides and let x=1/2,a=7/(7/2)=2
multiply (x+3) on both sides and let x=-3,b=1
2010-01-14 5:25 am
a/(2x-1) - b/(x+3) + 2/(2x^2+5x-3) = 9/(2x-1)(x+3)
a/(2x-1) - b/(x+3) + 2/(2x-1)(x+3) = 9/(2x-1)(x+3)
multiply (2x-1) on both sides and let x=1/2,a=7/(7/2)=2
multiply (x+3) on both sides and let x=-3,b=1
a/(2x-1) - b/(x+3) + 2/(2x-1)(x+3) = 9/(2x-1)(x+3)
multiply (2x-1) on both sides and let x=1/2,a=7/(7/2)=2
multiply (x+3) on both sides and let x=-3,b=1
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