Question

Alguém saberia fazer essa incógnita para me ajudar por favor....

Answer 1

Trocar o valor de g(x) pelo valor de x na função f(x):

$f(g(x))=5\left(\dfrac{3x+2}{x-1}\right)^2+\dfrac{3(3x+2)}{x-1}+2\\\\\\ f(g(x))=5\dfrac{(3x+2)^2}{(x-1)^2}+\dfrac{9x+6}{x-1}+2\\\\\\ f(g(x))=\dfrac{5(9x^2+12x+4)}{(x-1)^2}+\dfrac{9x+6}{x-1}+2\\\\\\ f(g(x))=\dfrac{45x^2+60x+20}{(x-1)^2}+\dfrac{9x+6}{x-1}+2\\\\\\ f(g(x))=\dfrac{45x^2+60x+20+(9x+6)(x-1)+2(x-1)^2}{(x-1)^2}\\\\\\ f(g(x))=\dfrac{45x^2+60x+20+9x^2-9x+6x-6+2(x^2-2x+1)}{(x-1)^2}\\\\\\ f(g(x))=\dfrac{45x^2+60x+20+9x^2-9x+6x-6+2x^2-4x+2}{(x-1)^2}$

$\boxed{\boxed{f(g(x))=\dfrac{56x^2+53x+16}{(x-1)^2}}}$

Espero ter ajudado.
Bons estudos!

Answer 2

             $f(x) = 5x^2+3x+2$
             $g(x) = \frac{3x+2}{x-1} \\ \\ f(g(x)) \\ \\ =5( \frac{3x+2}{x-1})^2 +3( \frac{3x+2}{x-1} )+2 \\ \\ = 5(\frac{9x^2+12x+4}{x^2-2x+1} )+ \frac{9x+6}{x-1} +2 \\ \\ = \frac{45x^2+60x+20}{x^2-2x+1} + \frac{9x+6}{x-1} +2 \\ \\ = \frac{45x^2+60x+20+(x-1)(9x+6)+2(x^2-2x+1)}{x^2-2x+1} \\ \\ = \frac{45x^2+60x+20+9x^2+6x-9x-6+2x^2-4x+2}{x^2-2x+1}$

             $f(g(x))= \frac{56x^2+53x+16}{x^2-2x+1}$