(2/x)+(3/(x+1))+(4/(x+2))+(5/(x+3))+(6/(x+4))=5


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Question Number 156063 by cortano last updated on 07/Oct/21

$\frac{2}{x} + \frac{3}{x + 1} + \frac{4}{x + 2} + \frac{5}{x + 3} + \frac{6}{x + 4} = 5$
Commented by john_santu last updated on 07/Oct/21

$x = {2, - 2 \pm \sqrt{\frac{15 \pm \sqrt{145}}{10}}}$
Commented by Tawa11 last updated on 07/Oct/21

$nice$
	Answered by ajfour last updated on 07/Oct/21

	$l e t \frac{4}{x + 2} = t$ $\Rightarrow x = \frac{4}{t} - 2$ $\frac{t}{2 - t} + \frac{3 t}{4 - t} + t + \frac{5 t}{4 + t} + \frac{3 t}{2 + t} = 5$ $\frac{2 t - 2}{2 - t} + \frac{4 t - 4}{4 - t} + t - 1 + \frac{4 t - 4}{4 + t} + \frac{2 t - 2}{2 + t} = 0$ $2 (\frac{4}{4 - t^{2}}) + 4 (\frac{8}{16 - t^{2}}) + 1 = 0$ $a p a r t f r o m t = 1 \Rightarrow x = 2$ $l e t 10 - t^{2} = z$ $\frac{8}{z - 6} + \frac{32}{z + 6} + 1 = 0$ $40 z - 144 = 36 - z^{2}$ $\Rightarrow z^{2} + 40 z - 180 = 0$ $z = - 20 \pm \sqrt{400 + 180}$ $z = - 20 \pm \sqrt{580}$ $x = \frac{4}{t} - 2 = \frac{\pm 4}{\sqrt{10 - z}} - 2$ $x = \frac{\pm 4}{\sqrt{30 \mp \sqrt{580}}} - 2$ $x = \frac{\pm 4 \sqrt{(30 \pm \sqrt{580})}}{\sqrt{320}} - 2$ $x = \pm (\sqrt{\frac{15 \pm \sqrt{145}}{10}}) - 2$ $& x_{1} = 2$
	Commented by Tawa11 last updated on 08/Oct/21

	$Great sir$
	Answered by Rasheed.Sindhi last updated on 08/Oct/21

	$\frac{2}{x} + \frac{3}{x + 1} + \frac{4}{x + 2} + \frac{5}{x + 3} + \frac{6}{x + 4} = 5$ $\frac{2}{x} - 1 + \frac{3}{x + 1} - 1 + \frac{4}{x + 2} - 1 + \frac{5}{x + 3} - 1 + \frac{6}{x + 4} - 1 = 0$ $\frac{2 - x}{x} + \frac{3 - x - 1}{x + 1} + \frac{4 - x - 2}{x + 2} + \frac{5 - x - 3}{x + 3} + \frac{6 - x - 4}{x + 4} = 0$ $\frac{2 - x}{x} + \frac{2 - x}{x + 1} + \frac{2 - x}{x + 2} + \frac{2 - x}{x + 3} + \frac{2 - x}{x + 4} = 0$ $(2 - x) (\frac{1}{x} + \frac{1}{x + 1} + \frac{1}{x + 2} + \frac{1}{x + 3} + \frac{1}{x + 4}) = 0$ $x = 2 ∣ \frac{1}{x} + \frac{1}{x + 1} + \frac{1}{x + 2} + \frac{1}{x + 3} + \frac{1}{x + 4} = 0$ $(\frac{1}{x} + \frac{1}{x + 4}) + (\frac{1}{x + 1} + \frac{1}{x + 3}) + \frac{1}{x + 2} . \frac{2 (x + 2)}{2 (x + 2)} = 0$ $\frac{2 x + 4}{x (x + 4)} + \frac{2 x + 4}{(x + 1) (x + 3)} + \frac{2 x + 4}{2 {(x + 2)}^{2}} = 0$ $\frac{1}{x (x + 4)} + \frac{1}{(x + 1) (x + 3)} + \frac{1}{2 {(x + 2)}^{2}} = 0 [∵ x \neq - 2]$ $\frac{1}{x^{2} + 4 x} + \frac{1}{x^{2} + 4 x + 3} + \frac{1}{2 (x^{2} + 4 x + 4)} = 0$ $L e t x^{2} + 4 x = y$ $\frac{1}{y} + \frac{1}{y + 3} + \frac{1}{2 (y + 4)} = 0$ $2 (y + 3) (y + 4) + 2 y (y + 4) + y (y + 3) = 0$ ${2 y}^{2} + 14 y + 24 + {2 y}^{2} + 8 y + y^{2} + 3 y = 0$ ${5 y}^{2} + 25 y + 24 = 0$ $y = \frac{- 25 \pm \sqrt{625 - 480}}{10}$ $x^{2} + 4 x = \frac{- 25 \pm \sqrt{145}}{10}$ ${10 x}^{2} + 40 x + 25 \mp \sqrt{145} = 0$ $x = \frac{- 40 \pm \sqrt{1600 - 1000 \pm 40 \sqrt{145}}}{20}$ $x = \frac{- 40 \pm 2 \sqrt{150 \pm 10 \sqrt{145}}}{20}$ $x = \frac{- 20 \pm \sqrt{150 \pm 10 \sqrt{145}}}{10}$ $= - 2 \pm \sqrt{\frac{15 \pm \sqrt{145}}{10}}, 2$
	Answered by Rasheed.Sindhi last updated on 08/Oct/21
	$N^(A) E^(A^(S I^(⋏^•^∣ ) ) E) R WAY (2/x)+(3/(x+1))+(4/(x+2))+(5/(x+3))+(6/(x+4))=5 x+2=y: (2/(y−2))+(3/(y−1))+(4/y)+(5/(y+1))+(6/(y+2))=5 (2/(y−2))−1+(3/(y−1))−1+(4/y)−1+(5/(y+1))−1+(6/(y+2))−1=0 ((2−y+2)/(y−2))+((3−y+1)/(y−1))+((4−y)/y)+((5−y−1)/(y+1))+((6−y−2)/(y+2))=0 ((4−y)/(y−2))+((4−y)/(y−1))+((4−y)/y)+((4−y)/(y+1))+((4−y)/(y+2))=0 (4−y)((1/(y−2))+(1/(y−1))+(1/y)+(1/(y+1))+(1/(y+2)))=0 { ((4−y=0⇒y=4⇒x+2=4⇒x=2)),(((1/(y−2))+(1/(y−1))+(1/y)+(1/(y+1))+(1/(y+2))=0)) :} ((1/(y−2))+(1/(y+2)))+((1/(y−1))+(1/(y+1)))+(1/y)=0 ((2y)/(y^2 −4))+((2y)/(y^2 −1))+(1/y).((2y)/(2y))=0 2y((1/(y^2 −4))+(1/(y^2 −1))+(1/(2y^2 )))=0 (1/(y^2 −4))+(1/(y^2 −1))+(1/(2y^2 ))=0 [∵y≠0] (1/(y^2 −4))+(1/(y^2 −1))=((−1)/(2y^2 )) ((y^2 −1+y^2 −4)/((y^2 −4)(y^2 −1)))=((−1)/(2y^2 )) 2y^2 (2y^2 −5)=−(y^2 −4)(y^2 −1) 4y^4 −10y^2 =−y^4 +5y^2 −4 5y^4 −15y^2 +4=0 y^2 =((15±(√(225−80)))/(10))=((15±(√(145)))/(10)) (x+2)^2 =((15±(√(145)))/(10)) x=−2±(√((15±(√(145)))/(10))) x=2 , −2±(√((15±(√(145)))/(10)))$
	$\overset{A}{N} E^{A^{\overset{\overset{\overset{∣}{∙}}{⋏}}{S I}} E} R WAY$ $\frac{2}{x} + \frac{3}{x + 1} + \frac{4}{x + 2} + \frac{5}{x + 3} + \frac{6}{x + 4} = 5$ $x + 2 = y :$ $\frac{2}{y - 2} + \frac{3}{y - 1} + \frac{4}{y} + \frac{5}{y + 1} + \frac{6}{y + 2} = 5$ $\frac{2}{y - 2} - 1 + \frac{3}{y - 1} - 1 + \frac{4}{y} - 1 + \frac{5}{y + 1} - 1 + \frac{6}{y + 2} - 1 = 0$ $\frac{2 - y + 2}{y - 2} + \frac{3 - y + 1}{y - 1} + \frac{4 - y}{y} + \frac{5 - y - 1}{y + 1} + \frac{6 - y - 2}{y + 2} = 0$ $\frac{4 - y}{y - 2} + \frac{4 - y}{y - 1} + \frac{4 - y}{y} + \frac{4 - y}{y + 1} + \frac{4 - y}{y + 2} = 0$ $(4 - y) (\frac{1}{y - 2} + \frac{1}{y - 1} + \frac{1}{y} + \frac{1}{y + 1} + \frac{1}{y + 2}) = 0$ ${\begin{cases} 4 - y = 0 \Rightarrow y = 4 \Rightarrow x + 2 = 4 \Rightarrow x = 2 \\ \frac{1}{y - 2} + \frac{1}{y - 1} + \frac{1}{y} + \frac{1}{y + 1} + \frac{1}{y + 2} = 0 \end{cases}$ $(\frac{1}{y - 2} + \frac{1}{y + 2}) + (\frac{1}{y - 1} + \frac{1}{y + 1}) + \frac{1}{y} = 0$ $\frac{2 y}{y^{2} - 4} + \frac{2 y}{y^{2} - 1} + \frac{1}{y} . \frac{2 y}{2 y} = 0$ $2 y (\frac{1}{y^{2} - 4} + \frac{1}{y^{2} - 1} + \frac{1}{{2 y}^{2}}) = 0$ $\frac{1}{y^{2} - 4} + \frac{1}{y^{2} - 1} + \frac{1}{{2 y}^{2}} = 0 [∵ y \neq 0]$ $\frac{1}{y^{2} - 4} + \frac{1}{y^{2} - 1} = \frac{- 1}{{2 y}^{2}}$ $\frac{y^{2} - 1 + y^{2} - 4}{(y^{2} - 4) (y^{2} - 1)} = \frac{- 1}{{2 y}^{2}}$ ${2 y}^{2} ({2 y}^{2} - 5) = - (y^{2} - 4) (y^{2} - 1)$ ${4 y}^{4} - {10 y}^{2} = - y^{4} + {5 y}^{2} - 4$ ${5 y}^{4} - {15 y}^{2} + 4 = 0$ $y^{2} = \frac{15 \pm \sqrt{225 - 80}}{10} = \frac{15 \pm \sqrt{145}}{10}$ ${(x + 2)}^{2} = \frac{15 \pm \sqrt{145}}{10}$ $x = - 2 \pm \sqrt{\frac{15 \pm \sqrt{145}}{10}}$ $x = 2, - 2 \pm \sqrt{\frac{15 \pm \sqrt{145}}{10}}$
	Commented by Tawa11 last updated on 08/Oct/21

	$Great sir$
	Commented by Rasheed.Sindhi last updated on 08/Oct/21

	$T hanks miss!$
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