Question

Se considera expresia:$(\frac{ x^{2}+5x+6 }{ x^{2} -3x-10} +2)· \frac{ x^{2} -25}{9 x^{2} -42x+49} unde x ∈ R\ \left \{ {{ -\frac{7}{3} ;-2;5 \right\} }} .$
Arati ca E(x)=$\frac{x+5}{3x-7} .$

Answer 1

   
[tex]\displaystyle\\ E(x)=\left(\frac{ x^{2}+5x+6 }{ x^{2} -3x-10} +2\right) \cdot \frac{ x^{2} -25}{9 x^{2} -42x+49}\\\\ \text{unde } x \in R\setminus\left \{ -\frac{7}{3} ;-2;5 \right\} \\\\ E(x)=\frac{x^{2}+5x+6 +2(x^{2} -3x-10)}{x^{2} -3x-10} \cdot \frac{ x^{2} -25}{9 x^{2} -42x+49}=\\\\ =\frac{x^{2}+5x+6 +2x^{2} -6x-20}{x^{2} -3x-10} \cdot \frac{ x^{2} -25}{9 x^{2} -42x+49}=\\\\ =\frac{3x^{2}-x-14 }{x^{2} -3x-10} \cdot \frac{ x^{2} -25}{9 x^{2} -42x+49}=[/tex]


[tex]\displaystyle\\ =\frac{3x^{2}+6x-7x-14 }{x^{2} +2x-5x-10} \cdot \frac{(x-5)(x+5)}{(3x)^{2} +2\cdot 3x\cdot (-7)+(-7)^2}=\\\\ =\frac{3x(x+2)-7(x+2)}{x(x +2)-5(x+2)} \cdot \frac{(x-5)(x+5)}{(3x)^{2} +2\cdot 3x\cdot (-7)+(-7)^2}=\\\\ =\frac{(x+2)(3x-7)}{(x +2)(x-5)} \cdot \frac{(x-5)(x+5)}{(3x-7)^{2}}=\\\\ =\frac{(x+2)(3x-7)}{(x +2)(x-5)} \cdot \frac{(x-5)(x+5)}{(3x-7)(3x-7)}=\\\\ = \frac{x+2}{x+2} \cdot \frac{x-5}{x-5} \cdot \frac{(3x-7)(x+5)}{(3x-7)(3x-7)}=\boxed{\bf \frac{x+5}{3x-7}}\\\\ \bf cctd [/tex]