+0  
 
0
771
1
avatar

simplify as far as possible  (x-2/(x+1))÷(1-(4x+7)/(x^2+4x+3))  

 May 26, 2015

Best Answer 

 #1
avatar+118723 
+10

$$\\\left(x-\frac{2}{(x+1)}\right)\div \left(1-\frac{4x+7}{x^2+4x+3} \right) \\\\
=\left(\frac{x(x+1)-2}{(x+1)}\right)\div \left(\frac{x^2+4x+3-(4x+7)}{x^2+4x+3} \right) \\\\
=\left(\frac{x^2+x-2}{(x+1)}\right)\div \left(\frac{x^2-4}{x^2+4x+3} \right) \\\\
=\frac{x^2+x-2}{(x+1)}\times \frac{x^2+4x+3} {x^2-4}\\\\
=\frac{(x+2)(x-1)}{(x+1)}\times \frac{(x+3)(x+1)} {(x-2)(x+2)} \\\\
=\frac{(x-1)}{1}\times \frac{(x+3)} {(x-2)} \\\\
=\frac{(x+1)(x+3)} {(x-2)} \\\\
=\frac{x^x+4x+3} {x-2} \\\\$$

 

If I haven't made any stupid mistakes then that will be ok :)

You need to check it!

 May 26, 2015
 #1
avatar+118723 
+10
Best Answer

$$\\\left(x-\frac{2}{(x+1)}\right)\div \left(1-\frac{4x+7}{x^2+4x+3} \right) \\\\
=\left(\frac{x(x+1)-2}{(x+1)}\right)\div \left(\frac{x^2+4x+3-(4x+7)}{x^2+4x+3} \right) \\\\
=\left(\frac{x^2+x-2}{(x+1)}\right)\div \left(\frac{x^2-4}{x^2+4x+3} \right) \\\\
=\frac{x^2+x-2}{(x+1)}\times \frac{x^2+4x+3} {x^2-4}\\\\
=\frac{(x+2)(x-1)}{(x+1)}\times \frac{(x+3)(x+1)} {(x-2)(x+2)} \\\\
=\frac{(x-1)}{1}\times \frac{(x+3)} {(x-2)} \\\\
=\frac{(x+1)(x+3)} {(x-2)} \\\\
=\frac{x^x+4x+3} {x-2} \\\\$$

 

If I haven't made any stupid mistakes then that will be ok :)

You need to check it!

Melody May 26, 2015

0 Online Users