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Question Number 162785 by john_santu last updated on 01/Jan/22

	Answered by Rasheed.Sindhi last updated on 02/Jan/22
	$((x^4 +2x+5)/(x^2 (x+1)))=(x−1)+((x^2 +2x+5)/(x^2 (x+1))) ((x^2 +2x+5)/(x^2 (x+1)))=(A_1 /x)+(A_2 /x^2 )+(B/(x+1)) A_1 x(x+1)+A_2 (x+1)+Bx^2 =x^2 +2x+5 x=0:A_2 (0+1)=0^2 +2.0+5=5 A_2 =5 x=−1:B(−1)^2 =(−1)^2 +2(−1)+5 B=4 A_1 x^2 +A_1 x+A_2 x+A_2 +Bx^2 =x^2 +2x+5 (A_1 +B)x^2 +(A_1 +A_2 )x+A_2 =x^2 +2x+5 A_1 +B=1 , A_1 +A_2 =2 , A_2 =5 A_1 +B=1⇒A_1 +4=1⇒A_1 =−3 A_1 +A_2 =2⇒−3+5=2 ✓ ((x^4 +2x+5)/(x^2 (x+1)))=(x−1)+((x^2 +2x+5)/(x^2 (x+1))) =x−(3/x)+(5/x^2 )+(4/(x+1))−1 Σ_(x∈{a,b,c,d}) ((x^4 +2x+5)/(x^2 (x+1)))=Σ_(x∈{a,b,c,d}) x−Σ_(x∈{a,b,c,d}) (3/x) +Σ_(x∈{a,b,c,d}) (5/x^2 )+Σ_(x∈{a,b,c,d}) +Σ_(x∈{a,b,c,d}) (4/(x+1)) Σ_(x∈{a,b,c,d}) ((x^4 +2x+5)/(x^2 (x+1)))=−1+(Σ_(x∈{a,b,c,d}) x)− 3(Σ_(x∈{a,b,c,d}) (1/x)) +5(Σ_(x∈{a,b,c,d}) (1/x^2 )) +4Σ_(x∈{a,b,c,d}) (1/(x+1)) =−1+(a+b+c+d) −3((1/a)+(1/b)+(1/c)+(1/d)) +5((1/a^2 )+(1/b^2 )+(1/c^2 )+(1/d^2 )) +4((1/(a+1))+(1/(b+1))+(1/(c+1))+(1/(d+1))) .... ...$
	$\frac{x^{4} + 2 x + 5}{x^{2} (x + 1)} = (x - 1) + \frac{x^{2} + 2 x + 5}{x^{2} (x + 1)}$ $\frac{x^{2} + 2 x + 5}{x^{2} (x + 1)} = \frac{A_{1}}{x} + \frac{A_{2}}{x^{2}} + \frac{B}{x + 1}$ $A_{1} x (x + 1) + A_{2} (x + 1) + {B x}^{2}$ $= x^{2} + 2 x + 5$ $x = 0 : A_{2} (0 + 1) = 0^{2} + 2.0 + 5 = 5$ $A_{2} = 5$ $x = - 1 : B {(- 1)}^{2} = {(- 1)}^{2} + 2 (- 1) + 5$ $B = 4$ $A_{1} x^{2} + A_{1} x + A_{2} x + A_{2} + {B x}^{2}$ $= x^{2} + 2 x + 5$ $(A_{1} + B) x^{2} + (A_{1} + A_{2}) x + A_{2}$ $= x^{2} + 2 x + 5$ $A_{1} + B = 1, A_{1} + A_{2} = 2, A_{2} = 5$ $A_{1} + B = 1 \Rightarrow A_{1} + 4 = 1 \Rightarrow A_{1} = - 3$ $A_{1} + A_{2} = 2 \Rightarrow - 3 + 5 = 2 ✓$ $\frac{x^{4} + 2 x + 5}{x^{2} (x + 1)} = (x - 1) + \frac{x^{2} + 2 x + 5}{x^{2} (x + 1)}$ $= x - \frac{3}{x} + \frac{5}{x^{2}} + \frac{4}{x + 1} - 1$ $\sum_{x \in {a, b, c, d}} \frac{x^{4} + 2 x + 5}{x^{2} (x + 1)} = \underset{x \in {a, b, c, d}}{Σ} x - \underset{x \in {a, b, c, d}}{Σ} \frac{3}{x}$ $+ \underset{x \in {a, b, c, d}}{Σ} \frac{5}{x^{2}} + \underset{x \in {a, b, c, d}}{Σ} + \underset{x \in {a, b, c, d}}{Σ} \frac{4}{x + 1}$ $\sum_{x \in {a, b, c, d}} \frac{x^{4} + 2 x + 5}{x^{2} (x + 1)} = - 1 + (\underset{x \in {a, b, c, d}}{Σ} x) - 3 (\underset{x \in {a, b, c, d}}{Σ} \frac{1}{x})$ $+ 5 (\underset{x \in {a, b, c, d}}{Σ} \frac{1}{x^{2}}) + 4 \underset{x \in {a, b, c, d}}{Σ} \frac{1}{x + 1}$ $= - 1 + (a + b + c + d)$ $- 3 (\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d})$ $+ 5 (\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}} + \frac{1}{d^{2}})$ $+ 4 (\frac{1}{a + 1} + \frac{1}{b + 1} + \frac{1}{c + 1} + \frac{1}{d + 1})$ $. . . .$ $. . .$
	Commented by Rasheed.Sindhi last updated on 03/Jan/22

	$T han k s miss tawa,$ $y o u r c o m p l i m e n t s m a t t e r s f o r m e!$
	Commented by Tawa11 last updated on 03/Jan/22

	$Weldone sir$
	Commented by Tawa11 last updated on 04/Jan/22

	$I usually learn from your solutions sir . God bless you more .$
	Commented by Rasheed.Sindhi last updated on 04/Jan/22

	$T hanks Miss! Iappreciate your contribution to$ $to forum & your encouraging others .$ $GOD BLESS YOU ALSO!$
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