Odpowiedź:
[tex]\huge\boxed{~~\dfrac{m+k}{n+k}=-2~~}[/tex]
Szczegółowe wyjaśnienie:
Korzystamy ze wzoru skróconego mnożenia:
[tex]V(x)=(x-1)^{2}\cdot (x+k)\\\\V(x)=(x^{2} -2x+1)\cdot (x+k)\\\\V(x)=x^{3}+kx^{2} -2x^{2} -2kx+x+k\\\\V(x)=x^{3}+(k-2)x^{2} +(1-2k)x+k[/tex]
[tex]V(x)=1\cdot x^{3}+\boxed{(k-2)}x^{2} +\boxed{\boxed{(1-2k)}}x+ \boxed{\boxed{\boxed{k}}} \\~\\~\land~~\\\\W(x)=1\cdot x^{3}+\boxed{m}x^{2} +\boxed{\boxed{n}}x+\boxed{\boxed{\boxed{-3}}}\\\\\Downarrow\\\\\huge\boxed{k=-3}\\\\m=k-2~~\land~~k=-3~~\Rightarrow~~\huge\boxed{m=-5}\\\\n=1-2k~~\land~~k=-3~~\Rightarrow~~\huge\boxed{m=7}[/tex]
[tex]\dfrac{m+k}{n+k}=?~~\land~~k=-3~~\land~~m=-5~~\land~~n=7\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Downarrow\\\\\dfrac{-5+(-3)}{7+(-3)} =\dfrac{-5-3}{7-3} =\dfrac{-8}{4} =\huge\boxed{-2}[/tex]
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Odpowiedź:
[tex]\huge\boxed{~~\dfrac{m+k}{n+k}=-2~~}[/tex]
Szczegółowe wyjaśnienie:
Równość wielomianów
Korzystamy ze wzoru skróconego mnożenia:
Obliczamy:
[tex]V(x)=(x-1)^{2}\cdot (x+k)\\\\V(x)=(x^{2} -2x+1)\cdot (x+k)\\\\V(x)=x^{3}+kx^{2} -2x^{2} -2kx+x+k\\\\V(x)=x^{3}+(k-2)x^{2} +(1-2k)x+k[/tex]
[tex]V(x)=1\cdot x^{3}+\boxed{(k-2)}x^{2} +\boxed{\boxed{(1-2k)}}x+ \boxed{\boxed{\boxed{k}}} \\~\\~\land~~\\\\W(x)=1\cdot x^{3}+\boxed{m}x^{2} +\boxed{\boxed{n}}x+\boxed{\boxed{\boxed{-3}}}\\\\\Downarrow\\\\\huge\boxed{k=-3}\\\\m=k-2~~\land~~k=-3~~\Rightarrow~~\huge\boxed{m=-5}\\\\n=1-2k~~\land~~k=-3~~\Rightarrow~~\huge\boxed{m=7}[/tex]
[tex]\dfrac{m+k}{n+k}=?~~\land~~k=-3~~\land~~m=-5~~\land~~n=7\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Downarrow\\\\\dfrac{-5+(-3)}{7+(-3)} =\dfrac{-5-3}{7-3} =\dfrac{-8}{4} =\huge\boxed{-2}[/tex]