What is the answer to sif(x)=((2x-1))/(2x+1) ?

The answer to sif(x)=((2x-1))/(2x+1) is

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sif(x)=((2x-1))/(2x+1)

s i f (x) = \frac{( 2 x - 1 )}{2 x + 1}

(s = 0, s \in R, x = \frac{1}{2}, x = \frac{1}{2}, f \in R f = 0)

Solution steps

s i f (x) = \frac{( 2 x - 1 )}{2 x + 1}

Multiply both sides by

2 x + 1

s i f (x) (2 x + 1) = \frac{2 x - 1}{2 x + 1} (2 x + 1)

Expand

i (s f (x) + 2 s x f (x)) = 2 x - 1

Rewrite

i (s f (x) + 2 s x f (x))

in standard complex form:

0 + (s f (x) + 2 s x f (x)) i

0 + (s f (x) + 2 s x f (x)) i = 2 x - 1

Rewrite

2 x - 1

in standard complex form:

(2 x - 1) + 0 i

0 + (s f (x) + 2 s x f (x)) i = (2 x - 1) + 0 i

Complex numbers can be equal only if their real and imaginary parts are equalRewrite as system of equations:

[0 = 2 x - 1 s f (x) + 2 s x f (x) = 0]

[0 = 2 x - 1 s f (x) + 2 s x f (x) = 0] : (s = 0, s \in R, x = \frac{1}{2}, x = \frac{1}{2}, f \in R f = 0)

(s = 0, s \in R, x = \frac{1}{2}, x = \frac{1}{2}, f \in R f = 0)

i (x) = e (2 x^{2} - 1)

z \cdot z + z = 3 - i

a + bi = - i

x + 2 i = 5 + y i

x^{2} + i x 3 + i = 0

What is the answer to sif(x)=((2x-1))/(2x+1) ?
The answer to sif(x)=((2x-1))/(2x+1) is