Resolva:
{[(-2/5)^2 × (0,2)^ -1] - 7} × (0,3)^3=
leimores:
2,-1 e o 3 são potencia
{016*5-7} X 0.027
QUE E IGUAL A -0.1674
Todas as respostas dadas estão corretas e bem explicadas. ^^
Soluções para a tarefa
Respondido por
1
Olá.
Vamos montar e logo após resolver a expressão.
![\Large\begin{array}{c}\mathsf{\left\{ \left[ \left( \dfrac{-2}{5}\right)^2\cdot\left(0,2)^{-1} \right]-7\right\}\cdot \left( 0,3\right)^3=}\\\\\\\end{array}](https://tex.z-dn.net/?f=%5CLarge%5Cbegin%7Barray%7D%7Bc%7D%5Cmathsf%7B%5Cleft%5C%7B+%5Cleft%5B+%5Cleft%28+%5Cdfrac%7B-2%7D%7B5%7D%5Cright%29%5E2%5Ccdot%5Cleft%280%2C2%29%5E%7B-1%7D+%5Cright%5D-7%5Cright%5C%7D%5Ccdot+%5Cleft%28+0%2C3%5Cright%29%5E3%3D%7D%5C%5C%5C%5C%5C%5C%5Cend%7Barray%7D "\Large\begin{array}{c}\mathsf{\left\{ \left[ \left( \dfrac{-2}{5}\right)^2\cdot\left(0,2)^{-1} \right]-7\right\}\cdot \left( 0,3\right)^3=}\\\\\\\end{array}")
Primeiro, vamos nos atentar em algumas coisas:
' Todo número elevado a expoente negativo se inverte. Ex.:
'' Vamos transformar tanto o 0,3 quanto o 0,2 em fração.
Vamos continuar os cálculos.
![\Large\begin{array}{l}\mathsf{\left\{ \left[ \left( \dfrac{-2}{5}\right)^2\cdot\left(0,2\right)^{-1} \right]-7\right\}\cdot \left( 0,3\right)^3=}\\\\\\
\mathsf{\left\{ \left[ \left( \dfrac{-2}{5}\right)^2\cdot\left(\dfrac{2}{10}\right)^{-1} \right]-7\right\}\cdot \left( \dfrac{3}{10}\right)^3=}\\\\\\
\mathsf{\left\{ \left[ \left( \dfrac{(-2)^2}{(5)^2}\right)\cdot\left(\dfrac{10}{2}\right) \right]-7\right\}\cdot \left( \dfrac{(3)^3}{(10)^3}\right)=}\\\\\\\end{array}](https://tex.z-dn.net/?f=%5CLarge%5Cbegin%7Barray%7D%7Bl%7D%5Cmathsf%7B%5Cleft%5C%7B+%5Cleft%5B+%5Cleft%28+%5Cdfrac%7B-2%7D%7B5%7D%5Cright%29%5E2%5Ccdot%5Cleft%280%2C2%5Cright%29%5E%7B-1%7D+%5Cright%5D-7%5Cright%5C%7D%5Ccdot+%5Cleft%28+0%2C3%5Cright%29%5E3%3D%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B%5Cleft%5C%7B+%5Cleft%5B+%5Cleft%28+%5Cdfrac%7B-2%7D%7B5%7D%5Cright%29%5E2%5Ccdot%5Cleft%28%5Cdfrac%7B2%7D%7B10%7D%5Cright%29%5E%7B-1%7D+%5Cright%5D-7%5Cright%5C%7D%5Ccdot+%5Cleft%28+%5Cdfrac%7B3%7D%7B10%7D%5Cright%29%5E3%3D%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B%5Cleft%5C%7B+%5Cleft%5B+%5Cleft%28+%5Cdfrac%7B%28-2%29%5E2%7D%7B%285%29%5E2%7D%5Cright%29%5Ccdot%5Cleft%28%5Cdfrac%7B10%7D%7B2%7D%5Cright%29+%5Cright%5D-7%5Cright%5C%7D%5Ccdot+%5Cleft%28+%5Cdfrac%7B%283%29%5E3%7D%7B%2810%29%5E3%7D%5Cright%29%3D%7D%5C%5C%5C%5C%5C%5C%5Cend%7Barray%7D "\Large\begin{array}{l}\mathsf{\left\{ \left[ \left( \dfrac{-2}{5}\right)^2\cdot\left(0,2\right)^{-1} \right]-7\right\}\cdot \left( 0,3\right)^3=}\\\\\\
\mathsf{\left\{ \left[ \left( \dfrac{-2}{5}\right)^2\cdot\left(\dfrac{2}{10}\right)^{-1} \right]-7\right\}\cdot \left( \dfrac{3}{10}\right)^3=}\\\\\\
\mathsf{\left\{ \left[ \left( \dfrac{(-2)^2}{(5)^2}\right)\cdot\left(\dfrac{10}{2}\right) \right]-7\right\}\cdot \left( \dfrac{(3)^3}{(10)^3}\right)=}\\\\\\\end{array}")
![\Large\begin{array}{l}\mathsf{\left\{ \left[ \left( \dfrac{4}{25}\right)\cdot\left(\dfrac{10}{2}\right) \right]-7\right\}\cdot \left( \dfrac{27}{1.000}\right)=}\\\\\\
\mathsf{\left\{ \left[~\dfrac{4\cdot10}{25\cdot2}~\right]-7\right\}\cdot \left( \dfrac{27}{1.000}\right)=}\\\\\\
\mathsf{\left\{ \left[~\dfrac{40}{50}~\right]-7\right\}\cdot \left( \dfrac{27}{1.000}\right)=}\\\\\\
\mathsf{\left\{ \dfrac{40-7\cdot50}{50}\right\}\cdot \left( \dfrac{27}{1.000}\right)=}\\\\\\\end{array}](https://tex.z-dn.net/?f=%5CLarge%5Cbegin%7Barray%7D%7Bl%7D%5Cmathsf%7B%5Cleft%5C%7B+%5Cleft%5B+%5Cleft%28+%5Cdfrac%7B4%7D%7B25%7D%5Cright%29%5Ccdot%5Cleft%28%5Cdfrac%7B10%7D%7B2%7D%5Cright%29+%5Cright%5D-7%5Cright%5C%7D%5Ccdot+%5Cleft%28+%5Cdfrac%7B27%7D%7B1.000%7D%5Cright%29%3D%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B%5Cleft%5C%7B+%5Cleft%5B%7E%5Cdfrac%7B4%5Ccdot10%7D%7B25%5Ccdot2%7D%7E%5Cright%5D-7%5Cright%5C%7D%5Ccdot+%5Cleft%28+%5Cdfrac%7B27%7D%7B1.000%7D%5Cright%29%3D%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B%5Cleft%5C%7B+%5Cleft%5B%7E%5Cdfrac%7B40%7D%7B50%7D%7E%5Cright%5D-7%5Cright%5C%7D%5Ccdot+%5Cleft%28+%5Cdfrac%7B27%7D%7B1.000%7D%5Cright%29%3D%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B%5Cleft%5C%7B+%5Cdfrac%7B40-7%5Ccdot50%7D%7B50%7D%5Cright%5C%7D%5Ccdot+%5Cleft%28+%5Cdfrac%7B27%7D%7B1.000%7D%5Cright%29%3D%7D%5C%5C%5C%5C%5C%5C%5Cend%7Barray%7D "\Large\begin{array}{l}\mathsf{\left\{ \left[ \left( \dfrac{4}{25}\right)\cdot\left(\dfrac{10}{2}\right) \right]-7\right\}\cdot \left( \dfrac{27}{1.000}\right)=}\\\\\\
\mathsf{\left\{ \left[~\dfrac{4\cdot10}{25\cdot2}~\right]-7\right\}\cdot \left( \dfrac{27}{1.000}\right)=}\\\\\\
\mathsf{\left\{ \left[~\dfrac{40}{50}~\right]-7\right\}\cdot \left( \dfrac{27}{1.000}\right)=}\\\\\\
\mathsf{\left\{ \dfrac{40-7\cdot50}{50}\right\}\cdot \left( \dfrac{27}{1.000}\right)=}\\\\\\\end{array}")
Podemos simplificar essa fração, dividindo-a por 10.
Qualquer dúvida, deixe nos comentários.
Bons estudos.
Vamos montar e logo após resolver a expressão.
Primeiro, vamos nos atentar em algumas coisas:
' Todo número elevado a expoente negativo se inverte. Ex.:
'' Vamos transformar tanto o 0,3 quanto o 0,2 em fração.
Vamos continuar os cálculos.
Podemos simplificar essa fração, dividindo-a por 10.
Qualquer dúvida, deixe nos comentários.
Bons estudos.
Respondido por
3
{ (( - 2 )² x ( 0, 2 ) - ¹ ] - 7 } x ( 0, 3 ) ³
----
5
{ 4 ( 2 ) - ¹ } - 7 x 0.027
---- x -----
25 10
{ { 4 10 } - 7 } x 0,027
----- x -----
25 2
{ 40 - 7 } ⇒ 4 - 7 x 0.027
---- ---
50 5
4 - 35 x 0.027
-------------------
5
- 31 0,027
------- x
5
- 0,837
------- ou - 0,1674
5
----
5
{ 4 ( 2 ) - ¹ } - 7 x 0.027
---- x -----
25 10
{ { 4 10 } - 7 } x 0,027
----- x -----
25 2
{ 40 - 7 } ⇒ 4 - 7 x 0.027
---- ---
50 5
4 - 35 x 0.027
-------------------
5
- 31 0,027
------- x
5
- 0,837
------- ou - 0,1674
5
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