Matemática, perguntado por lorenadomingues606, 1 ano atrás

(1/4:9/32-10/9.2/5).³ raiz 81/256 pvf me ajudem

Soluções para a tarefa

Respondido por TesrX
51
Olá.

\mathsf{\left(\dfrac{1}{4}:\dfrac{9}{32}-\dfrac{10}{9}\cdot\dfrac{2}{5}\right)^3\cdot\sqrt{\dfrac{81}{256}}=}\\\\\\
\mathsf{\left(\dfrac{1}{4}\cdot\dfrac{32}{9}-\dfrac{10\cdot2}{9\cdot5}\right)^3\cdot\dfrac{9}{16}=}\\\\\\
\mathsf{\left(\dfrac{1\cdot32}{4\cdot9}-\dfrac{20}{45}\right)^3\cdot\dfrac{9}{16}=}\\\\\\
\mathsf{\left(\dfrac{32}{36}-\dfrac{20}{45}\right)^3\cdot\dfrac{9}{16}=}\\\\\\
\mathsf{\left(\dfrac{45\cdot32-36\cdot20}{36\cdot45}\right)^3\cdot\dfrac{9}{16}=}

\mathsf{\left(\dfrac{1.440-720}{1.620}\right)^3\cdot\dfrac{9}{16}=}\\\\\\
\mathsf{\left(\dfrac{720}{1.620}\right)^3\cdot\dfrac{9}{16}=}\\\\\\
\mathsf{\left(\left[\dfrac{720}{1.620}\right]^{:180}\right)^3\cdot\dfrac{9}{16}=}\\\\\\
\mathsf{\left(\dfrac{4}{9}\right)^3\cdot\dfrac{9}{16}=}\\\\\\\textsf{Igualando~pra~"cortar"...}\\\\
\mathsf{\dfrac{4^3}{9^3}\cdot\dfrac{9^1}{4^2}=}\\\\\\
\mathsf{\dfrac{4^3\cdot9^1}{9^3\cdot4^2}=}\\\\\\
\mathsf{\dfrac{4^1\cdot1}{9^2\cdot1}=}\\\\\\
\mathsf{\dfrac{4}{81}}\\\\\mathsf{0,049383...}

Qualquer dúvida, deixe nos comentários.
Bons estudos.
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