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Mi aiutereste con le espressioni n.12? Grazie.. non ci sono i risultati! 

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2

√((3/2 - 1/6 - 4/5) + (3/8·(4/9) + 1)·(9/2)·(16/5) - 40/3)=

=√(8/15 + (1/6 + 1)·(9/2)·(16/5) - 40/3)=

=√(8/15 + 7/6·(9/2)·(16/5) - 40/3)=

=√(8/15 + 84/5 - 40/3)=

=√4= 2

-------------------------------------------------

√(5/3·(1/3 + 1/5 - 1/15)·(1/7 + 1/4 + 1/2))=

=√(5/3·(7/15)·(25/28))=

=√(25/36) = 5/6



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a)

$\small \sqrt{\left(\dfrac{3}{2}-\dfrac{1}{6}-\dfrac{4}{5}\right)+\left(\dfrac{\cancel3^1}{\cancel8_2}×\dfrac{\cancel4^1}{\cancel9_3}+1\right)×\dfrac{9}{\cancel2_1}×\dfrac{\cancel{16}^8}{5}-\dfrac{40}{3}}=$

$\small =\sqrt{\left(\dfrac{45-5-24}{30}\right)+\left(\dfrac{1}{2}×\dfrac{1}{3}+1\right)×9×\dfrac{8}{5}-\dfrac{40}{3}}=$

$\small =\sqrt{\dfrac{16}{30}+\left(\dfrac{1}{6}+1\right)×\dfrac{72}{5}-\dfrac{40}{3}}=$

$\small =\sqrt{\dfrac{\cancel{16}^8}{\cancel{30}_{15}}+\left(\dfrac{1+6}{6}\right)×\dfrac{72}{5}-\dfrac{40}{3}}=$

$\small =\sqrt{\dfrac{8}{15}+\dfrac{7}{\cancel6_1}×\dfrac{\cancel{72}^{12}}{5}-\dfrac{40}{3}}=$

$\small =\sqrt{\dfrac{8}{15}+7×\dfrac{12}{5}-\dfrac{40}{3}}=$

$\small =\sqrt{\dfrac{8}{15}+\dfrac{84}{5}-\dfrac{40}{3}}=$

$\small =\sqrt{\dfrac{8+252-240}{15}}=$

$\small =\sqrt{\dfrac{\cancel{60}^4}{\cancel{15}_1}}= \sqrt4 = 2$

 

=======================================================

b)

$\small \sqrt{\dfrac{5}{3}×\left(\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{15}\right)×\left(\dfrac{1}{7}+\dfrac{1}{4}+\dfrac{1}{2}\right)} = $

$\small =\sqrt{\dfrac{5}{3}×\left(\dfrac{5+3-1}{15}\right)×\left(\dfrac{4+7+14}{28}\right)} = $

$\small =\sqrt{\dfrac{\cancel5^1}{3}×\dfrac{\cancel7^1}{\cancel{15}_3}×\dfrac{25}{\cancel{28}_4}} = $

$\small =\sqrt{\dfrac{1}{3}×\dfrac{1}{3}×\dfrac{25}{4}} = $

$\small =\sqrt{\dfrac{25}{36}} = \dfrac{5}{6}$



0

Ciao ti mando in allegato la lettera a

17374813674348811350343483054667

 

Però ho sbagliato il 2 calcolo ho messo un 94/5 al posto di un 84/5 scusami lo rimetto qui sotto

17374817694498238380467362293981



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