Question

plz solve above question​

Answer 1

Given that , $\sf \dfrac{1}{p} + \dfrac{1}{q} + \dfrac{1}{x} = \dfrac{1}{p + q + x}$.

Need To Find : Value of x ?

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀Given that ,

$\dashrightarrow \sf \dfrac{1}{p} + \dfrac{1}{q} + \dfrac{1}{x} = \dfrac{1}{p + q + x} \:$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Solving \: the \: Given \::}}$

$\dashrightarrow \sf \dfrac{1}{p} + \dfrac{1}{q} + \dfrac{1}{x} = \dfrac{1}{p + q + x} \:\\\\ \dashrightarrow \sf \dfrac{1}{p} + \dfrac{1}{q} + \dfrac{1}{x} - \dfrac{1}{p + q + x} = 0 \:\\\\ \dashrightarrow \sf \bigg[ \dfrac{1}{p} + \dfrac{1}{q}\bigg] +\bigg[ \dfrac{1}{x} - \dfrac{1}{p + q + x} \bigg] = 0 \:\\\\ \dashrightarrow \sf \bigg[ \dfrac{p + q }{pq}\bigg] +\bigg[ \dfrac{(x + p + q )- x }{x ( p + q + x ) } \bigg] = 0 \:\\\\\dashrightarrow \sf \bigg[ \dfrac{p + q }{pq}\bigg] +\bigg[ \dfrac{x + p + q - x }{x ( p + q + x ) } \bigg] = 0 \:\\\\\dashrightarrow \sf \bigg[ \dfrac{p + q }{pq}\bigg] +\bigg[ \dfrac{x + p + q - x }{x ( p + q + x ) } \bigg] = 0 \:\\\\ \dashrightarrow \sf \bigg[ \dfrac{p + q }{pq}\bigg] +\bigg[ \dfrac{ p + q }{x ( p + q + x ) } \bigg] = 0 \:\\\\ \dashrightarrow \sf ( p + q ) \bigg[ \dfrac{1}{pq} + \dfrac{ 1 }{x ( p + q + x ) } \bigg] = 0 \:\\\\ \dashrightarrow \sf ( p + q ) \bigg[ \dfrac{ x^2 + xp + xq + pq }{ (pq) (x) ( p + q + x ) } \bigg] = 0 \:\\\\\dashrightarrow \sf ( p + q ) \bigg[ \dfrac{ x^2 + xq + xp + pq }{ (pq) (x) ( p + q + x ) } \bigg] = 0 \:\\\\ \dashrightarrow \sf ( p + q ) \bigg[ \dfrac{ x( x + q ) + p ( x + q ) }{ (pq) (x) ( p + q + x ) } \bigg] = 0 \:\\\\\dashrightarrow \sf ( p + q ) \bigg[ \dfrac{ ( x + p ) ( x + q ) }{ (pq) (x) ( p + q + x ) } \bigg] = 0 \:\\\\ \dashrightarrow \sf \bigg[ \dfrac{ ( x + p ) ( x + q ) }{ (pq) (x) ( p + q + x ) } \bigg] = \dfrac{0}{( p + q )} \:\\\\ \dashrightarrow \sf \bigg[ \dfrac{ ( x + p ) ( x + q ) }{ (pq) (x) ( p + q + x ) } \bigg] = 0 \:\\\\ \dashrightarrow \sf ( x + p ) ( x + q ) = 0 \times (pq) (x) ( p + q + x ) \:\\\\ \dashrightarrow \sf ( x + p ) ( x + q ) = 0 \:\\\\ \dashrightarrow \sf \:x\:=\:-p \:\:or\:\: x \:=\:-q \:$

$\dashrightarrow \:\underline {\boxed{\purple {\pmb{\frak{\:\:x\:=\:-p \:\:or\:\: \:-q \:}}}}}\:\:\bigstar \:\:$

$\qquad \therefore \underline {\sf Hence, The \:value \:of \: x \:can \: be \:\pmb{\bf{ - p \: or \:-q \:}}.}$

Answer 2

Answer:

$x = \frac{1}{2 - \frac{1}{2 - \frac{1}{2 - x} } } \\ = > x = \frac{1}{2 - \frac{1}{ \frac{2(2 - x) - 1}{2 - x} } } \\ = > x = \frac{1}{2 - \frac{2 - x}{4 - 2x - 1 } } \\ = > x = \frac{1}{ \frac{2(3 - 2x) - (2 - x)}{3 - 2x} } \\ = > x = \frac{3 - 2x}{6 - 4x - 2 + x} \\ = > x = \frac{3 - 2x}{4 - 3x } \\ = > 4x - 3 {x}^{2} = 3 - 2x \\ = > 6x - 3 {x}^{2} = 3$