Question

determine o valor numérico das seguintes expressões algébricas



b) m³·$m{3}-2 · (m-1)^{2}+ 5mm-2, para = \frac{1}{2}$



d)$\frac{a^{2} +b+c}{-a^{2-b^{2-c^{2} } } } ^{2}$ para a = $\frac{1}{2}$, b= 2 e c= 1


me ajudem eu n consigo fazer essas duas

Answer 1

Resposta:

d) $\frac{169}{441}$

Explicação passo a passo:

d) $(\frac{a^2 + b + c}{-a^2 - b^2 -c^2})^2$ = $(\frac{(\frac{1}{2})^2 + 2 + 1}{-(\frac{1}{2}) ^2 - 2^2 -1^2})^2$= $(\frac{\frac{1}{4} + 2 + 1}{-\frac{1}{4} - 4 -})^2$= $(\frac{\frac{1 + 8 + 4}{4}}{\frac{-1 - 16 - 4}{4} })^2$ = $(\frac{1 + 8 + 4}{-1 - 16 - 4})^2$ = $(-\frac{13}{21})^2$ =$\frac{169}{441}$

Espero ter ajudado ; )

Answer 2

Resposta:

$b)$ $\frac{1}{8}$

$d)$ $\frac{169}{441}$

Explicação passo a passo:

$b)$ $m^{3} -2(m-1)^{2} +5m-2$

$m=\frac{1}{2}$

$(\frac{1}{2} )^{3} -2(\frac{1}{2} -1)^{2} +5(\frac{1}{2}) -2$

$\frac{1^{3}}{2^{3}} -2(\frac{1}{2} -1(\frac{2}{2} ))^{2} +\frac{5}{2} -2(\frac{2}{2} )$

$\frac{1}{8} -2(\frac{1}{2} -\frac{2}{2} )^{2} +\frac{5}{2} -\frac{4}{2}$

$\frac{1}{8} -2(\frac{1-2}{2} )^{2} +\frac{5-4}{2}$

$\frac{1}{8} -2(-\frac{1}{2} )^{2} +\frac{1}{2}$

$\frac{1}{8} -2(-\frac{1}{2} )(-\frac{1}{2} ) +\frac{1}{2}$

$\frac{1}{8} -2(\frac{1}{4} ) +\frac{1}{2}$

$\frac{1}{8} -\frac{2}{4} +\frac{1}{2}$

$\frac{1}{8} -\frac{2:2}{4:2} +\frac{1}{2}$

$\frac{1}{8} -\frac{1}{2} +\frac{1}{2}$

$\frac{1}{8}$

$.............................$

$d)$ $(\frac{a^{2}+b+c }{-a^{2}-b^{2}-c^{2} })^{2}$

$a=\frac{1}{2} ,b=2,c=1$

$(\frac{(\frac{1}{2} )^{2}+2+1 }{-(\frac{1}{2} )^{2}-2^{2}-1^{2} })^{2}$

$(\frac{\frac{1^{2}}{2^{2}} +3 }{-\frac{1^{2}}{2^{2}} -4-1 })^{2}$

$(\frac{\frac{1}{4} +3 }{-\frac{1}{4} -5 })^{2}$

$(\frac{\frac{1}{4} +3(\frac{4}{4} ) }{-\frac{1}{4} -5(\frac{4}{4} ) })^{2}$

$(\frac{\frac{1}{4} +\frac{12}{4} }{-\frac{1}{4} -\frac{20}{4} })^{2}$

$(\frac{\frac{1+12}{4} }{-\frac{1+20}{4}})^{2}$

$(\frac{\frac{13}{4} }{-\frac{21}{4}})^{2}$

$\frac{(\frac{13}{4})^{2} }{(-\frac{21}{4})^{2} }$

$\frac{\frac{13^{2}}{4^{2}} }{\frac{(-21)^{2}}{4^{2}} }$

$\frac{\frac{169}{16} }{\frac{441}{16} }$

$\frac{169}{16} :\frac{441}{16}$

$\frac{169}{16} .\frac{16}{441}$

$\frac{169}{441} .\frac{16}{16}$

$\frac{169}{441} .1$

$\frac{169}{441}$