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andreyandreev [35.5K]
3 years ago
15

X^ 3 +x^ 2 +4x/x^ 2 +x-2into partialfractions​

Mathematics
1 answer:
lawyer [7]3 years ago
6 0

Answer:

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

Step-by-step explanation:

x + (6x/ x^2 + 2x - x -2)

x + (6x/ (x + 2) X (x - 1))

(6x/ (x + 2) X (x - 1))

(A/ x+2) + (B/ x-1)

(6x/ (x + 2) X (x - 1)) = (A/ x+2) + (B/ x-1)

6x = Ax + Bx - A + 2B

6x = (A+B)x + (-A+2b)

{0 = -A+2B

{6 = A+B

(A,B) = (4, 2)

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

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

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Fully detailed and equipped information process right below. The answer for this question is "25" (simplified form).

\mathbf{2^{28} \times \Bigg(\dfrac{5^{- 2}}{2^3} \Bigg)^4 \times \Bigg(5^{- 5} \times 2^8 \times 1 \Bigg)^{- 2}}

\mathbf{2^{28} \times \Bigg(\dfrac{5^{- 2}}{2^3} \Bigg)^4 \times \Bigg(2^8 \times 1 \times \dfrac{1}{5^5} \Bigg)^{- 2}}

\mathbf{2^{28} \times \Bigg(\dfrac{5^{- 2}}{2^3} \Bigg)^4 \times \Bigg(1 \times \dfrac{2^8 \times 1}{5^5} \Bigg)^{- 2}}

\mathbf{2^{28} \times \Bigg(\dfrac{5^{- 2}}{2^3} \Bigg)^4 \times \Bigg(1 \times \dfrac{256}{3125} \Bigg)^{- 2}}

\mathbf{2^{28} \times \Bigg(\dfrac{5^{- 2}}{2^3} \Bigg)^4 \times \Bigg(\dfrac{3125}{256} \Bigg)^2}

\mathbf{2^{28} \times \Bigg(\dfrac{5^{- 2}}{2^3} \Bigg)^4 \times \dfrac{3125^2}{256^2}}

\mathbf{2^{28} \times \Bigg(\dfrac{1}{2^3 \times 5^2} \Bigg)^4 \times \dfrac{3125^2}{256^2}}

\mathbf{2^{28} \times \Bigg(\dfrac{1^4}{(2^3 \times 5^2)^4} \Bigg) \times \dfrac{3125^2}{256^2}}

\mathbf{2^{28} \times \Bigg(\dfrac{1^4}{(5^2)^4 \times (2^3)^4} \Bigg) \times \dfrac{3125^2}{256^2}}

\mathbf{2^{28} \times \dfrac{1}{2^{12} \times 5^8} \times \dfrac{3125^2}{256^2}}

\mathbf{\dfrac{3125^2 \times 1 \times 2^{28}}{256^2 \times 5^8 \times 2^{12}}}

\mathbf{\dfrac{2^{(28 - 12)} \times 3125^2}{256^2 \times 5^8}}

\mathbf{\dfrac{2^{16} \times 3125^2}{256^2 \times 5^8}}

\mathbf{\dfrac{2^{16} \times (5^5)^2}{(2^8)^2 \times 5^8}}

\mathbf{\dfrac{2^{16} \times 5^{10}}{2^{16} \times 5^8}}

\mathbf{\dfrac{2^{16} \times 5^{10 - 8}}{2^{16}}}

\mathbf{\dfrac{5^{10 - 8}}{1}}

\mathbf{\therefore \quad 5^2}

\mathbf{\therefore \quad 25}

\boxed{\mathbf{\underline{\therefore \quad Final \: \: Answer \: \: is: \: 25}}}

Hope it helps.
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