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\(\text{I'm gonna type everything in }\LaTeX\)

\(\text{First Question:}\\ \quad \frac{1}{x}+\frac{1}{1-x}\\=\frac{1-x}{x(1-x)}+\frac{x}{x(1-x)}\\=\frac{1-x+x}{x-x^2}\\=\frac{1}{x-x^2}\)

\(\boxed{\color{red}\therefore \frac{1}{x}+\frac{1}{1-x}=\frac{1}{x-x^2}}\)

\(\text{Second Question:}\\ \quad 1+\frac{1}{x+1}\\=\frac{x+1}{x+1}+\frac{1}{x+1}\\=\frac{x+2}{x+1}\)

\(\boxed{\color{red}\therefore 1+\frac{1}{x+1}\neq \frac{2}{x+1}}\)

\(\text{Third Question:}\\\quad x+\frac{1}{x}\\=\frac{x^2}{x}+\frac{1}{x}\\=\frac{x^2+1}{x}\)

\(\boxed{\color{red}\therefore x+\frac{1}{x}=\frac{x^2+1}x}\)

\(\text{Last Question:}\\\quad\frac{1}{x-1}-\frac{1}{x+1}\\=\frac{x+1}{x^2-1}-\frac{x-1}{x^2-1}\\=\frac{x-x+1+1}{x^2-1}\\=\frac{2}{x^2-1}\)

\(\boxed{\color{red}\therefore\frac{1}{x-1}-\frac{1}{x+1}\neq \frac{2x}{1-x^2}}\)

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Is 1/x+1/1-x = 1/x-x^2 ?

 

Nope, 1/x+1/1-x = (1-x)/(x+1)

 

Is 1+ 1/x+1 = 2/x+1 ?

 

Nope, 1 + 1/(x+1) = (x+1)/(x+1) + 1/(x+1) = (x+2)/(x+1)

 

Is x+ 1/x = x^2+1/x ?

 

Yes, x + 1/x = x^2/x + 1/x = (x^2+1)/x

 

Is 1/x-1 - 1/x+1 = 2x/1-x^2 ?

 

Nope, 1/(x-1) - 1/(x+1) = (x+1)/(x-1) - (x-1)/(x+1) = ((x+1)-(x-1))/(x^2-1) = 2/(x^2-1)

 

A helpful way to check your answer is to graph the equation (that way you can instantly visually see if they equal each other).