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(x+1)/(2x-2)-(x-1)/(2x+2)-(4x)/(x^2-1)+(x^2+1)/(x^2-1)=(x-1)/(x+1)

Arman  Apr 3, 2015

Best Answer 

 #2
avatar+92805 
+10

 

I made an error is my first post so I have done it again

 

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

 

$$\\LHS=\frac{(x+1)}{2(x-1)}-\frac{(x-1)}{2(x+1)}+\frac{(x^2+1)}{(x^2-1)}-\frac{(4x)}{(x^2-1)}\\\\
LHS=\frac{(x+1)^2}{2(x-1)(x+1)}-\frac{(x-1)^2}{2(x+1)(x-1)}+\frac{(x^2+1)-4x}{(x^2-1)}\\\\
LHS=\frac{(x+1)^2-(x-1)^2}{2(x-1)(x+1)}+\frac{x^2+1-4x}{(x-1)(x+1)}\\\\
LHS=\frac{(x^2+2x+1)-(x^2-2x+1)}{2(x-1)(x+1)}+\frac{x^2-4x+1}{(x-1)(x+1)}\\\\
LHS=\frac{4x}{2(x-1)(x+1)}+\frac{x^2-4x+1}{(x-1)(x+1)}\\\\
LHS=\frac{2x}{(x-1)(x+1)}+\frac{x^2-4x+1}{(x-1)(x+1)}\\\\
LHS=\frac{2x+x^2-4x+1}{(x-1)(x+1)}\\\\
LHS=\frac{x^2-2x+1}{(x-1)(x+1)}\\\\
LHS=\frac{(x-1)(x-1)}{(x-1)(x+1)}\\\\
LHS=\frac{x-1}{x+1}\\\\
LHS =RHS$$

Melody  Apr 4, 2015
 #2
avatar+92805 
+10
Best Answer

 

I made an error is my first post so I have done it again

 

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

 

$$\\LHS=\frac{(x+1)}{2(x-1)}-\frac{(x-1)}{2(x+1)}+\frac{(x^2+1)}{(x^2-1)}-\frac{(4x)}{(x^2-1)}\\\\
LHS=\frac{(x+1)^2}{2(x-1)(x+1)}-\frac{(x-1)^2}{2(x+1)(x-1)}+\frac{(x^2+1)-4x}{(x^2-1)}\\\\
LHS=\frac{(x+1)^2-(x-1)^2}{2(x-1)(x+1)}+\frac{x^2+1-4x}{(x-1)(x+1)}\\\\
LHS=\frac{(x^2+2x+1)-(x^2-2x+1)}{2(x-1)(x+1)}+\frac{x^2-4x+1}{(x-1)(x+1)}\\\\
LHS=\frac{4x}{2(x-1)(x+1)}+\frac{x^2-4x+1}{(x-1)(x+1)}\\\\
LHS=\frac{2x}{(x-1)(x+1)}+\frac{x^2-4x+1}{(x-1)(x+1)}\\\\
LHS=\frac{2x+x^2-4x+1}{(x-1)(x+1)}\\\\
LHS=\frac{x^2-2x+1}{(x-1)(x+1)}\\\\
LHS=\frac{(x-1)(x-1)}{(x-1)(x+1)}\\\\
LHS=\frac{x-1}{x+1}\\\\
LHS =RHS$$

Melody  Apr 4, 2015

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