Question

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Answer 1

Resposta:

A resolução das expressões algébricas são:

c)$\frac{8\sqrt{2}}{3}$.

d)0.

e)$-\frac{8}{9}$.

f)-0,25.

g)$-\frac{65}{63}$.

h)0,5.

Por favor, acompanhar a Explicação.

Explicação passo a passo:

Vamos proceder aos cálculos solicitados:

Exercício c)

$\sqrt{\frac{a^{2}+ax}{m}}\\a=8\\x=10\\m=9\\\\ \sqrt{\frac{8^{2}+8.10}{9}}=\sqrt{\frac{64+80}{9}}=\sqrt{\frac{128}{9}}=\sqrt{\frac{2^{7}}{3^{2}}}=\sqrt{\frac{2^{6}.2^{1}}{3^{2}}}=\frac{2^{3}}{3}.\sqrt{2} =\frac{8}{3}\sqrt{2}=\frac{8\sqrt{2}}{3}$

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Exercício d)

3(x² - y²) - 10·(x + y)·(x - y) = 3(x² - y²) - 10·(x² - y²) = (x² - y²)·(3 - 10) = (x² - y²)·(-7) = -7·(x² - y²)

$-7.(x^{2} - y^{2})\\x=-2\\y=-2\\\\-7.[(-2)^{2} - (-2)^{2}]=-7.[(4)-(4)]=-7.[4-4]=-7.[0]=-7.0=0$

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Exercício e)

(a - b)² - c² = (a - b)² - (c)² = [(a - b) + c]·[(a - b) - c] = [(a - b + c)·(a - b - c)] = (a - b + c)·(a - b - c)

$(a-b+c).(a-b-c)\\a=\frac{2}{3}\\b=1\\c=-1\\\\(\frac{2}{3}-1+(-1)).(\frac{2}{3}-1-(-1))=(\frac{2}{3}-1-1).(\frac{2}{3}-1+1)=(\frac{2}{3}-2).(\frac{2}{3}+0)=(\frac{2-6}{3}).(\frac{2}{3})=(\frac{-4}{3}).(\frac{2}{3})=(\frac{-4.2}{3.3})=\frac{-8}{9}=-\frac{8}{9}$

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Exercício f)

$\frac{1-x^{2}}{xy+1}\\x=0,5\\y=-8\\\\\frac{1-0,5^{2}}{0,5.(-8)+1}=\frac{1-0,25}{-4+1}=\frac{0,75}{-3}=-\frac{0,75}{3}=-0,25$

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Exercício g)

$\frac{x^{3}-y^{3}}{x^{3}+y{3}}\\x=\frac{1}{2}\\y=-2\\\\ \frac{(\frac{1}{2})^{3}-(-2)^{3}}{(\frac{1}{2})^{3}+(-2)^{{3}}}=\frac{(\frac{1}{8})-(-8)}{(\frac{1}{8})+(-8)}=\frac{\frac{1}{8}+8}{\frac{1}{8}-8}=\frac{\frac{1+64}{8}}{\frac{1-64}{8}}=\frac{\frac{65}{8}}{\frac{-63}{8}}=\frac{65}{-63}=-\frac{65}{63}$

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Exercício h)

$\frac{y+\frac{1}{x}}{x+\frac{1}{y}}\\x=10\\y=5\\\\ \frac{5+\frac{1}{10}}{10+\frac{1}{5}}=\frac{\frac{50+1}{10}}{\frac{50+1}{5}}=\frac{\frac{51}{10}}{\frac{51}{5}}=\frac{51}{10}.\frac{5}{51}=\frac{5}{10}=\frac{1}{2}=0,5$