Question

Se dau numerele:
n=3⁴⁰:(3³x3⁴)x(3²+3x25):42 si m=411014-411x2³x5³-2³x7²:28.Calculati m la n,n la m,m+n,m·n.
 

Answer 1

n = 3^33 ·3·28:42 = 2·3^33
m = 411014 - 411000 - 14= 0
m^n = 0       n^m =1        m+n =2·3^33      m·n =0

Answer 2

$n= 3^{40} :( 3^{3} *3^{4} )*( 3^{2} +3*25):42=$$3^{40}:3^{7}*(9+75):42=3 ^{33} *84:42=3 ^{33}*2$

$m=411014-411* 2^{3}* 5^{3}- 2^{3} * 7^{2} :28=$$411014-411*8*125- \frac{2^{3} * 7^{2}}{2 ^{2}*7 } =411014-411000-14=14-14=0$

$\m^{n} =0 ^{(3 ^{33}*2) } =0$

$n^{m} =( {3 ^{33}*2 })^0 =1$

$m+n=0+3 ^{33} *2=3 ^{33} *2$

$m-n=0-3 ^{33} *28=-3 ^{33} *2$