Answers: 1
![g(x)=\dfrac{3x^3}{x^2+x-4}\\\\a)\\g(-2)=\dfrac{3(-2)^3}{(-2)^2+(-2)-4}=\dfrac{3(-8)}{4-2-4}=\dfrac{-24}{-2}=12\\\\b)\\g(5x)=\dfrac{3(5x)^3}{(5x)^2+(5x)-4}=\dfrac{3(125x^3)}{25x^2+5x-4}=\dfrac{375x^3}{25x^2+5x-4}\\\\c)\\g(8-4b)=\dfrac{3(8-4b)^3}{(8-4b)^2+(8-4b)-4}=\dfrac{3[8^3-3\cdot8^2\cdot4b+3\cdot8\cdot(4b)^2-(4b)^3]}{64-64b+16b^2+8-4b-4}=\\\\=\dfrac{3(512-768b+384b^2-64b^3)}{16b^2-68b+68}=\dfrac{3\cdot4(128-192b+96b^2-16b^3)}{4(4b^2-17b+17)}=\\\\=\dfrac{384-576b+288b^2-48b^3}{4b^2-17b+17}](/tpl/images/0738/1340/779a6.png)
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