Question

Binômio de newton (qual é a propriedade usada?)

Answer 1

$\displaystyle \text m+\text n = 4 \\\\ \text m^5-{5\choose 1}\text m^4\text n+{5\choose 2}\text m^3\text n^2-{5\choose 3}\text m^2\text n^3+{5\choose 4}\text m.\text n^4-\text n^5=32$

Sabemos que :

$\displaystyle (\text m-\text n)^{\text p}=\sum_{\text k = 0}^{\text p}{\text p\choose \text k}(-1)^{\text k}.\text m^{(\text p-\text k)}.\text n^{\text k} \\\\ \underline{\text{repare que se fizermos p = 5, teremos}}: \\\\ (\text m-\text n)^{5}={5\choose0}(-1)^{0}.\text m^{(5-0)}.\text n^{0}+{5\choose 1}(-1)^{1}.\text m^{5-1}.\text n^{1} +...+{5\choose 5}(-1)^{5}\text m^{(5-5)}.\text n^{5}$

$\displaystyle (\text m-\text n)^5=\text m^5-{5\choose 1}\text m^4\text n+{5\choose 2}\text m^3\text n^2-{5\choose 3}\text m^2\text n^3+{5\choose 4}\text m.\text n^4-\text n^5$

Então concluímos temos :

$\displaystyle \text m+\text n = 4 \\\\ (\text m-\text n)^5=32 \to \text m-2=\sqrt[5]{32} = 2 \\\\\\ \underline{\text{o sistema fica assim}}: \\\\ \text m+\text n=4 \\\\ \underline {\text m-\text n=2 \ + }\\\\ 2\text m =6 \\\\\ \huge\boxed{\text m = 3\ }\checkmark$

letra d