+0

# Help me please..

0
282
5
+44

Perform the following and simplify:

1. $${\frac{{\frac{\left(\left({\frac{{\mathtt{x}}}{\left({{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}}\right){\mathtt{\,-\,}}\left({\frac{{\mathtt{3}}}{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}}\right)\right)}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{3}}\right)}}}{\left({{\mathtt{x}}}^{{\mathtt{3}}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}}$$

2. $$\left({\frac{{\mathtt{1}}}{\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{2}}\right)}}{\mathtt{\,-\,}}{\frac{{\mathtt{2}}}{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{3}}}{{\mathtt{x}}}}\right){\mathtt{\,\times\,}}\left({\frac{\left(\left({{\mathtt{x}}}^{{\mathtt{2}}}\right){\mathtt{\,\small\textbf+\,}}{\mathtt{x}}\right)}{\left({{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{3}}\right)}}\right)$$

3. $${\frac{\left({\frac{{\mathtt{x}}}{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{y}}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{y}}}{\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{y}}\right)}}\right){\mathtt{\,\times\,}}\left({\frac{\left({{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{x}}{\mathtt{\,\times\,}}{\mathtt{y}}\right)}{\left({{\mathtt{x}}}^{{\mathtt{4}}}{\mathtt{\,-\,}}{{\mathtt{y}}}^{{\mathtt{4}}}\right)}}\right)}{\left({\frac{{\mathtt{x}}}{\left(\left({{\mathtt{x}}}^{{\mathtt{2}}}\right){\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\times\,}}{\mathtt{y}}{\mathtt{\,\small\textbf+\,}}\left({{\mathtt{y}}}^{{\mathtt{2}}}\right)\right)}}\right)}}$$

Phanolo  Jul 29, 2015

### Best Answer

#4
+18829
+13

$$3.\\\\ \small{\text{ \dfrac{ \left( \dfrac{x}{x+y}+\dfrac{y}{x-y}\right) * \left( \dfrac{ x^2-xy }{ x^4-y^4 } \right) } { \dfrac{ x } { x^2+2xy+y^2 } } }}$$

$$\small{\text{ \begin{array}{rcl} &=& \left( \dfrac{x}{x+y}+\dfrac{y}{x-y}\right)* \left( \dfrac{ x^2-xy }{ x^4-y^4 } \right) * \left( \dfrac { x^2+2xy+y^2 } { x } \right) \\\\ &=& \left( \dfrac{x}{x+y}+\dfrac{y}{x-y}\right)* \left( \dfrac{ x(x-y) }{ x^4-y^4 } \right) * \left( \dfrac { (x+y)^2 } { x } \right) \\\\ &=& \left( \dfrac{x}{x+y}+\dfrac{y}{x-y}\right)* \left( \dfrac{ (x-y)(x+y)^2 }{ x^4-y^4 } \right)\\\\ &=& \left( \dfrac{x}{x+y}+\dfrac{y}{x-y}\right)* \left( \dfrac{ (x-y)(x+y)(x+y) }{ x^4-y^4 } \right)\\\\ &=& \left( \dfrac{x}{x+y}+\dfrac{y}{x-y}\right)* \left( \dfrac{ (x^2-y^2)(x+y) }{ x^4-y^4 } \right)\\\\ &=& \left( \dfrac{x}{x+y}+\dfrac{y}{x-y}\right)* \left( \dfrac{ (x^2-y^2)(x+y) }{ (x^2-y^2)(x^2+y^2) } \right)\\\\ &=& \left( \dfrac{x}{x+y}+\dfrac{y}{x-y}\right)* \left( \dfrac{ x+y }{ x^2+y^2 } \right)\\\\ &=& \left( \dfrac{x(x-y)+y(x+y)}{(x-y)(x+y)}\right)* \left( \dfrac{ x+y }{ x^2+y^2 } \right)\\\\ &=& \left( \dfrac{ x^2+y^2 }{(x-y)(x+y)}\right)* \left( \dfrac{ x+y }{ x^2+y^2 } \right)\\\\ &\mathbf{=} & \mathbf{ \dfrac{ 1 }{x-y} } \end{array} }}$$

heureka  Jul 29, 2015
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### 5+0 Answers

#1
+91450
+10

1.     ((x/(x^2-1)) - (3/(x+1)))/(2x^2-x-3)/(x^3-1)

YOU WILL NEED TO CHECK MY ANSWER, I COULD EASILY HAVE MADE ONE OR MORE CARELESS ERRORS.

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

Melody  Jul 29, 2015
#2
+91450
+10

2.

(1/(x-2) - 2/(x+1) + 3/x) * (((x^2)+x)/(x^2+x-3)))

$$\left({\frac{{\mathtt{1}}}{\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{2}}\right)}}{\mathtt{\,-\,}}{\frac{{\mathtt{2}}}{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{3}}}{{\mathtt{x}}}}\right){\mathtt{\,\times\,}}\left({\frac{\left(\left({{\mathtt{x}}}^{{\mathtt{2}}}\right){\mathtt{\,\small\textbf+\,}}{\mathtt{x}}\right)}{\left({{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{3}}\right)}}\right)$$

$$\\\left(\frac{1}{(x-2)} -\frac{ 2}{(x+1)} + \frac{3}{x}\right) \times \frac{x(x+1)}{x^2+x-3}\\\\ =\left(\frac{x(x+1)}{x(x+1)(x-2)} -\frac{ 2x(x-2)}{x(x-2)(x+1)} + \frac{3(x+1)(x-2)}{x(x+1)(x-2)}\right) \times \frac{x(x+1)}{x^2+x-3}\\\\ =\frac{x(x+1)-2x(x-2)+3(x+1)(x-2)}{x(x+1)(x-2)} \times \frac{x(x+1)}{x^2+x-3}\\\\ =\frac{x^2+x-2x^2+4x+3(x^2-x-2)}{(x-2)} \times \frac{1}{x^2+x-3}\\\\ =\frac{x^2+x-2x^2+4x+3x^2-3x-6}{(x-2)} \times \frac{1}{x^2+x-3}\\\\ =\frac{2x^2+2x-6}{(x-2)} \times \frac{1}{x^2+x-3}\\\\ =\frac{2(x^2+x-3)}{(x-2)} \times \frac{1}{x^2+x-3}\\\\ =\frac{2}{x-2} \\\\$$

Again, you need to check my working although this one looks more promising than the first one that I did :)

Melody  Jul 29, 2015
#3
+91450
+10

$${\frac{\left({\frac{{\mathtt{x}}}{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{y}}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{y}}}{\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{y}}\right)}}\right){\mathtt{\,\times\,}}\left({\frac{\left({{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{x}}{\mathtt{\,\times\,}}{\mathtt{y}}\right)}{\left({{\mathtt{x}}}^{{\mathtt{4}}}{\mathtt{\,-\,}}{{\mathtt{y}}}^{{\mathtt{4}}}\right)}}\right)}{\left({\frac{{\mathtt{x}}}{\left(\left({{\mathtt{x}}}^{{\mathtt{2}}}\right){\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\times\,}}{\mathtt{y}}{\mathtt{\,\small\textbf+\,}}\left({{\mathtt{y}}}^{{\mathtt{2}}}\right)\right)}}\right)}}$$

$$\\\left(\left(\frac{x}{(x+y)}+\frac{y}{(x-y)}\right)*\left(\frac{(x^2-xy)}{(x^4-y^4)}\right)\right)\div\left(\frac{x}{(x^2)+2xy+(y^2)}\right)\\\\ =\left(\left(\frac{x(x-y)}{(x+y)(x-y)}+\frac{y(x+y)}{(x-y)(x+y)}\right)*\left(\frac{x(x-y)}{(x^2)^2-(y^2)^2}\right)\right)\times\left(\frac{(x^2)+2xy+(y^2)}{x}\right)\\\\ =\left(\left(\frac{x^2-xy+xy+y^2}{(x+y)(x-y)}\right)*\left(\frac{x(x-y)}{(x^2)^2-(y^2)^2}\right)\right)\times\left(\frac{(x+y)^2}{x}\right)\\\\ =\left(\left(\frac{x^2+y^2}{(x+y)}\right)*\left(\frac{x}{(x^2-y^2)(x^2+y^2)}\right)\right)\times\left(\frac{(x+y)^2}{x}\right)\\\\ =\frac{x^2+y^2}{1}\times\frac{1}{(x^2-y^2)(x^2+y^2)}\times\frac{x+y}{1}\\\\ =\frac{1}{1}\times\frac{1}{(x^2-y^2)}\times\frac{x+y}{1}\\\\ =\frac{1}{(x-y)(x+y)}\times\frac{x+y}{1}\\\\ =\frac{1}{x-y}\\\\$$

Melody  Jul 29, 2015
#4
+18829
+13
Best Answer

$$3.\\\\ \small{\text{ \dfrac{ \left( \dfrac{x}{x+y}+\dfrac{y}{x-y}\right) * \left( \dfrac{ x^2-xy }{ x^4-y^4 } \right) } { \dfrac{ x } { x^2+2xy+y^2 } } }}$$

$$\small{\text{ \begin{array}{rcl} &=& \left( \dfrac{x}{x+y}+\dfrac{y}{x-y}\right)* \left( \dfrac{ x^2-xy }{ x^4-y^4 } \right) * \left( \dfrac { x^2+2xy+y^2 } { x } \right) \\\\ &=& \left( \dfrac{x}{x+y}+\dfrac{y}{x-y}\right)* \left( \dfrac{ x(x-y) }{ x^4-y^4 } \right) * \left( \dfrac { (x+y)^2 } { x } \right) \\\\ &=& \left( \dfrac{x}{x+y}+\dfrac{y}{x-y}\right)* \left( \dfrac{ (x-y)(x+y)^2 }{ x^4-y^4 } \right)\\\\ &=& \left( \dfrac{x}{x+y}+\dfrac{y}{x-y}\right)* \left( \dfrac{ (x-y)(x+y)(x+y) }{ x^4-y^4 } \right)\\\\ &=& \left( \dfrac{x}{x+y}+\dfrac{y}{x-y}\right)* \left( \dfrac{ (x^2-y^2)(x+y) }{ x^4-y^4 } \right)\\\\ &=& \left( \dfrac{x}{x+y}+\dfrac{y}{x-y}\right)* \left( \dfrac{ (x^2-y^2)(x+y) }{ (x^2-y^2)(x^2+y^2) } \right)\\\\ &=& \left( \dfrac{x}{x+y}+\dfrac{y}{x-y}\right)* \left( \dfrac{ x+y }{ x^2+y^2 } \right)\\\\ &=& \left( \dfrac{x(x-y)+y(x+y)}{(x-y)(x+y)}\right)* \left( \dfrac{ x+y }{ x^2+y^2 } \right)\\\\ &=& \left( \dfrac{ x^2+y^2 }{(x-y)(x+y)}\right)* \left( \dfrac{ x+y }{ x^2+y^2 } \right)\\\\ &\mathbf{=} & \mathbf{ \dfrac{ 1 }{x-y} } \end{array} }}$$

heureka  Jul 29, 2015
#5
+44
+8

Thank you all! Those 3 items are the only things that I had a hard time solving

Phanolo  Jul 29, 2015

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