Question

A professora de Maria Eduarda escreveu no quadro a seguinte função:Determine o valor numérico da expressão ​

Answer 1

Explicação passo-a-passo:

• $\sf f(3)=\dfrac{(\sqrt{3\cdot3-5})\cdot(3+2)}{(3-1)}$

$\sf f(3)=\dfrac{(\sqrt{9-5})\cdot5}{2}$

$\sf f(3)=\dfrac{\sqrt{4}\cdot5}{2}$

$\sf f(3)=\dfrac{2\cdot5}{2}$

$\sf f(3)=\dfrac{10}{2}$

$\sf f(3)=5$

• $\sf f(2)=\dfrac{(\sqrt{3\cdot2-5})\cdot(2+2)}{(2-1)}$

$\sf f(2)=\dfrac{(\sqrt{6-5})\cdot4}{1}$

$\sf f(2)=\sqrt{1}\cdot4$

$\sf f(2)=1\cdot4$

$\sf f(2)=4$

• $\sf f(7)=\dfrac{(\sqrt{3\cdot7-5})\cdot(7+2)}{(7-1)}$

$\sf f(7)=\dfrac{(\sqrt{21-5})\cdot9}{6}$

$\sf f(7)=\dfrac{\sqrt{16}\cdot5}{6}$

$\sf f(7)=\dfrac{4\cdot9}{6}$

$\sf f(7)=\dfrac{36}{6}$

$\sf f(7)=6$

• $\sf \dfrac{f(3)-f(2)}{f(7)}=\dfrac{5-4}{6}$

$\sf \red{\dfrac{f(3)-f(2)}{f(7)}=\dfrac{1}{6}}$

Answer 2

Resposta:

1/6

Explicação passo-a-passo:

$\displaystyle\\f(x)=\frac{(\sqrt{3x-5} )(x+2)}{x-1} \\\\f(2)=\frac{(\sqrt{3.2-5} )(2+2)}{2-1} =\frac{(\sqrt{6-5} )(4)}{1} =\sqrt{1}.4=4 \\\\f(3)=\frac{(\sqrt{3.3-5} )(3+2)}{3-1}=\frac{(\sqrt{9-5} )(5)}{2}=\frac{(\sqrt{4} )(5)}{2}=\frac{(2)(5)}{2}=5\\\\f(7)=\frac{(\sqrt{3.7-5} )(7+2)}{7-1}=\frac{(\sqrt{21-5} )(9)}{6}=\frac{(\sqrt{16} )(3)}{2}=\frac{(4)(3)}{2}=6\\\\\\\frac{f(3)-f(2)}{f(7)} =\frac{5-4}{6} =\frac{1}{6}$