2-x-3-x-1-4-x-2-5-x-3-6-x-4-5-

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

\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!