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$$\left({\frac{{\mathtt{2}}}{\left({{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}}\right)}}\right){\mathtt{\,\small\textbf+\,}}\left({\frac{{\mathtt{5}}}{\left({{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right)}}\right) = \left({\frac{{\mathtt{6}}}{\left({{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}\right)}}\right) \Rightarrow {\mathtt{x}} = -{\mathtt{6}}$$

I need step by step solving this task..

 Jul 2, 2015

Best Answer 

 #1
avatar+118587 
+10

$$\begin{array}{rll}
\frac{2}{x^2-5x+6}+\frac{5}{x^2-4x+3}&=&\frac{6}{x^2-3x+2}\\\\
\frac{2}{(x-3)(x-2)}+\frac{5}{(x-3)(x-1)}&=&\frac{6}{(x-2)(x-1)}\\\\
(x-3)(x-2)(x-1)\left(\frac{2}{(x-3)(x-2)}+\frac{5}{(x-3)(x-1)}\right)&=&(x-3)(x-2)(x-1)\left(\frac{6}{(x-2)(x-1)}\right)\\\\
2(x-1)+5(x-2)&=&6(x-3)\\\\
2x-2+5x-10&=&6x-18\\\\
7x-12&=&6x-18\\\\
1x&=&-6\\\\
x&=&-6
\end{array}$$

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 Jul 2, 2015
 #1
avatar+118587 
+10
Best Answer

$$\begin{array}{rll}
\frac{2}{x^2-5x+6}+\frac{5}{x^2-4x+3}&=&\frac{6}{x^2-3x+2}\\\\
\frac{2}{(x-3)(x-2)}+\frac{5}{(x-3)(x-1)}&=&\frac{6}{(x-2)(x-1)}\\\\
(x-3)(x-2)(x-1)\left(\frac{2}{(x-3)(x-2)}+\frac{5}{(x-3)(x-1)}\right)&=&(x-3)(x-2)(x-1)\left(\frac{6}{(x-2)(x-1)}\right)\\\\
2(x-1)+5(x-2)&=&6(x-3)\\\\
2x-2+5x-10&=&6x-18\\\\
7x-12&=&6x-18\\\\
1x&=&-6\\\\
x&=&-6
\end{array}$$

Melody Jul 2, 2015

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