Richiesta risoluzione espressione

Non c'è nulla da risolvere: né enigmi, né equazioni, né problemi.
Si tratta di un'espressione esclusivamente numerica che si VALUTA (non si risolve!) per semplificazioni successive.
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A) "(-2/7)^-4 * (-2/7)^5 : (-2/7)^3" = (- 2/7)^(5 - 4 - 3) = (- 2/7)^(- 2) = (7/2)^2 = 49/4
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B) "(35/34)^-2 * (5/17)^2" = (5/17)^2/(35/34)^2 = ((5/17)/(35/34))^2 = (2/7)^2 = 4/49
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C) "[(-2/7)^-4 * (-2/7)^5 : (-2/7)^3]^3" = ((7/2)^2)^3 = (7/2)^6
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D) "[(35/34)^-2 * (5/17)^2]^2" = ((2/7)^2)^2 = (2/7)^4
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E) "[(-2/7)^-4 * (-2/7)^5 : (-2/7)^3]^3 * [(35/34)^-2 * (5/17)^2]^2" =
= ((7/2)^6)*(2/7)^4 = (7^6/2^6)*2^4/7^4 = 7^(6 - 4)/2^(6 - 4) = (7/2)^2
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F) "[(-2/7)^-4 * (-2/7)^5 : (-2/7)^3]^3 * [(35/34)^-2 * (5/17)^2]^2 * (7/4)^-2" =
= (7/2)^2/(7/4)^2 = (7^2/2^2)/(7^2/2^4) = 2^4/2^2 = 2^(4 - 2) = 4

 

((- 2/7)^(-4)·(- 2/7)^5/(- 2/7)^3)^3·((35/34)^(-2)·(5/17)^2)^2·(7/4)^(-2)=

=(7^4/2^4·(- 2^5/7^5)/(- 2^3/7^3))^3·(2^2·17^2/(5^2·7^2)·(5^2/17^2))^2·(2^4/7^2)=

=((- 2/7)/(- 2^3/7^3))^3·(2^2/7^2)^2·(2^4/7^2)=

=(7^2/2^2)^3·(2^2/7^2)^2·(2^4/7^2)=

=7^6/2^6·(2^4/7^4)·(2^4/7^2)=

=2^2 = 4

$\left[\left(-{2\over7}\right)^{-4}·\left(-{2\over7}\right)^5÷\left(-{2\over7}\right)^3\right]^3·\left[\left({35\over34}\right)^{-2}·\left({5\over17}\right)^2\right]^2·\left({7\over4}\right)^{-2}=$

$=\left[\left(-{2\over7}\right)^{-4+5-3}\right]^3 · \left[\left({34\over35}\right)^2 · \left({5\over17}\right)^2\right]^2 · \left({4\over7}\right)^2=$

$=\left[\left(-{2\over7}\right)^{-2}\right]^3 · \left[\left({34\over35}\right)^2 · \left({5\over17}\right)^2\right]^2 · \left({4\over7}\right)^2=$

$=\left(-{2\over7}\right)^{-2·3} · \left[\left({\cancel{34}^2\over\cancel{35}_7}\right)^2 · \left({\cancel5^1\over\cancel{17}_1}\right)^2\right]^2 · \left({4\over7}\right)^2=$

$=\left(-{2\over7}\right)^{-6} · \left[\left({2\over7}\right)^2 · \left({1\over1}\right)^2\right]^2 · \left({4\over7}\right)^2=$

$=\left(-{2\over7}\right)^{-6} · \left[\left({2\over7}\right)^2 ·1\right]^2 · \left({4\over7}\right)^2=$

$=\left(-{2\over7}\right)^{-6} ·\left({2\over7}\right)^{2·2} · \left({4\over7}\right)^2=$

$=\left(-{2\over7}\right)^{-6} ·\left({2\over7}\right)^4 · \left({4\over7}\right)^2=$

$=\left(-{2\over7}\right)^{-6+4} · \left({4\over7}\right)^2=$

$=\left(-{2\over7}\right)^{-2} · \left({4\over7}\right)^2=$

$=\left(-{7\over2}\right)^2 · \left({4\over7}\right)^2=$

$=\left(-{\cancel7^1\over\cancel2_1}\right)^2 · \left({\cancel4^2\over\cancel7_1}\right)^2=$

$=\left(-{1\over1}\right)^2 · \left({2\over1}\right)^2=$

$=1·4=$

$=4$

 

 

 

 

scusa ma a matita mi vien più veloce

ciao