7 – Calcule a seguinte expressão matemática:
3 . ( – 3/4 )^-2 + 6 . ( 3^-1/4 ) – 4
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7 . ( – 3/4 )^-1 + 2
3 . ( – 3/4 )^-2 + 6 . ( 3^-1/4 ) – 4
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7 . ( – 3/4 )^-1 + 2
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Para resolver esta expressão matemática usaremos a seguinte hierarquia de operações :D :
[tex] \textsf{ \boxed{ \begin{array}{ccc}\dfrac{\sf{3\cdot \left(-\dfrac{3}{4}\right)^{-2}+6\cdot\dfrac{3^{-1}}{4}-4}}{7\cdot \left(-\dfrac{3}{4}\right)^{-1}+2} \end{array}}}[/tex]
Aplicamos as propriedades nas potências para obter a expressão:
[tex] \textsf{ \boxed{ \begin{array}{ccc}\dfrac{\sf{3\cdot -\dfrac{3^{-2}}{4^{-2}}+6\cdot\dfrac{\dfrac{1}{3}}{4}-4}}{7\cdot -\dfrac{3^{-1}}{4^{-1}}+2} \end{array}}}[/tex]
[tex] \textsf{ \boxed{ \begin{array}{ccc}\dfrac{\sf{3\cdot -\dfrac{\dfrac{1}{3^2}}{\dfrac{1}{4^2}}+6\cdot\dfrac{\dfrac{1}{3}}{4}-4}}{7\cdot \dfrac{\dfrac{1}{3}}{\dfrac{1}{4}}+2} \end{array}}}[/tex]
[tex] \textsf{ \boxed{ \begin{array}{ccc}\dfrac{\sf{3\cdot -\dfrac{\dfrac{1}{9}}{\dfrac{1}{16}}+6\cdot\dfrac{\dfrac{1}{3}}{4}-4}}{7\cdot -\dfrac{\dfrac{1}{3}}{\dfrac{1}{4}}+2} \end{array}}}[/tex]
Efetuamos as divisões de frações se não as fizermos não podemos efetuar as multiplicações:
[tex] \textsf{ \boxed{ \begin{array}{ccc}\dfrac{\sf{3\cdot -\dfrac{16}{9}+6\cdot\dfrac{1}{12}-4}}{7\cdot -\dfrac{4}{3}+2} \end{array}}}[/tex]
[tex] \textsf{ \boxed{ \begin{array}{ccc}\dfrac{\sf{-\dfrac{16}{3}+\dfrac{1}{2}-4}}{7\cdot -\dfrac{28}{3}+2} \end{array}}}[/tex]
[tex] \textsf{ \boxed{ \begin{array}{ccc}\dfrac{\sf{-\dfrac{16}{3}-\dfrac{7}{2}}}{-\dfrac{28}{3}+2} \end{array}}}[/tex]
[tex] \textsf{ \boxed{ \begin{array}{ccc}\dfrac{\sf{-\dfrac{11}{6}}}{\dfrac{22}{3}} \end{array}}}[/tex]
E o resultado é:
[tex] \textsf{ \boxed{ \begin{array}{ccc}-\sf \dfrac{1}{4} =-\dfrac{3}{12}=-\dfrac{33}{132} \end{array}}}~~\blue \checkmark[/tex]
Dúvidas? Comente =)
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Resposta:
[tex]\textsf{Segue a resposta abaixo}[/tex]
Explicação passo-a-passo:
[tex] \mathsf{\left[\dfrac{3\cdot\left(-\dfrac{4}{3}\right)^2+6\cdot\left(\dfrac{1}{3}\cdot\dfrac{1}{4}\right)-4}{7\cdot\left(-\dfrac{4}{3}\right)+2}\right]=\left[\dfrac{3\cdot\left(-\dfrac{4}{3}\right)^2+6\cdot\left(\dfrac{1}{12}\right)-4}{-\dfrac{28}{3}+2}\right]}[/tex]
[tex] \mathsf{ \left[\dfrac{\diagup\!\!\!\!3\cdot\left(\dfrac{16}{\diagup\!\!\!\!3\cdot3}\right)+\diagup\!\!\!\!6\cdot\left(\dfrac{1}{\diagup\!\!\!\!6\cdot2}\right)-4}{\dfrac{(-28)+(2\cdot3)}{3}}\right]=\left[\dfrac{\dfrac{16}{3}+\dfrac{1}{2}-4}{-\dfrac{22}{3}}\right] }[/tex]
[tex] \mathsf{ \left[\dfrac{\dfrac{(16\cdot2)+3}{3\cdot2}-4}{-\dfrac{22}{3}}\right]=\left[\dfrac{\dfrac{35}{6}-4}{-\dfrac{22}{3}}\right] }[/tex]
[tex] \mathsf{ \left[\dfrac{\dfrac{35-(4\cdot6)}{6}}{-\dfrac{22}{3}}\right] =\left[\dfrac{\dfrac{11}{6}}{-\dfrac{22}{3}}\right]}[/tex]
[tex] \mathsf{ \left(\dfrac{~\diagup\!\!\!\!\!\!11}{2\cdot\diagup\!\!\!\!3}\right)\cdot\left(\dfrac{\diagup\!\!\!\!3}{~\diagup\!\!\!\!\!\!11\cdot(-2)}\right)=-\dfrac{1}{4}}[/tex]
[tex]\boxed{\boxed{\mathsf{\left[\dfrac{3\cdot\left(-\dfrac{3}{4}\right)^{-2}+6\cdot\left(\dfrac{3^{-1}}{4}\right)-4}{7\cdot\left(-\dfrac{3}{4}\right)^{-1}+2}\right]=-\dfrac{1}{4}}} }[/tex]