Question

Solve for x
$x = \dfrac{1}{2-\dfrac{1}{2-\dfrac{1}{2-x} } } \: \: \: , (x \neq 2)$

Answer 1

Step-by-step explanation:

Given :-

$\sf x = \dfrac{1}{2-\dfrac{1}{2-\dfrac{1}{2-x} } }$

To Find:-

• The value of x

Solution:-

$\sf x = \dfrac{1}{2-\dfrac{1}{2-\dfrac{1}{2-x} } }$

$\sf \dashrightarrow x = \dfrac{1}{2 - \dfrac{1}{\dfrac{2(2 - x) - 1}{2-x} } }$

$\sf \dashrightarrow x = \dfrac{1}{2 - \dfrac{1}{\dfrac{4 - 2x- 1}{2-x} } }$

$\sf \dashrightarrow x = \dfrac{1}{2 - \dfrac{2 - x}{{3 - 2x}} }$

$\sf \dashrightarrow x = \dfrac{1}{\dfrac{2(3 - 2x) - 2 - x}{{3 - 2x}} }$

$\sf \dashrightarrow x = \dfrac{1}{\dfrac{6 - 4x - 2 - x}{{3 - 2x}} }$

$\sf \dashrightarrow x = \dfrac{3 - 2x}{4 - 3x}$

$\sf \dashrightarrow x(4 - 3x) = {3 - 2x}$

$\sf \dashrightarrow 4x - 3 {x}^{2} = {3 - 2x}$

$\sf \dashrightarrow 3 {x}^{2} - 4x + 3 - 2x = 0$

$\sf \dashrightarrow 3 {x}^{2} - 6x + 3 = 0$

$\sf \dashrightarrow {x}^{2} - 2x + 1 = 0$

$\sf \dashrightarrow {(x - 1)}^{2} = 0$

$\sf \dashrightarrow (x - 1)(x - 1) = 0$

$\underline{ \boxed{\sf \purple{\dashrightarrow \: \: x \: = \: 1 \: \: , \: \: 1}}}$