Matematică, întrebare adresată de Utilizator anonim, 9 ani în urmă

Calculati:a)2x2^2x2^3+10;b)3^61:(3^4)^15+(5^2)^3:5^2^0;c)7^2:{17^2:289+2x[(2^3x3)^15:(2x2^43x3^15)+1^100]};d)5^140:25^70+8^32:16^24+100^25:10^48-2^2x5^2;e)[3^100:3^10+(9^5x3^14)^5:27^10+(4^30:4^29-1^20)^102:81^3]:(2^10:2^9+1)^90.Va rog frumos!


Kyoko: ce inseamna ^ ?
Utilizator anonim: este ridicarea la putere

Răspunsuri la întrebare

Răspuns de renatemambouko
9
a)2x2^2x2^3+10=2^6+10=64+10=74
b)3^61:(3^4)^15+(5^2)^3:5^2^0=
3^61:3^60+5^6:5^1=3+5^5=3+3125=3128
c)7^2:{17^2:289+2x[(2^3x3)^15:(2x2^43x3^15)+1^100]}=
=
7^2:{17^2:17^2+2x[(2^45x3^15):(2^44x3^15)+1]}=
=7^2:[1+2x(2+1)]=7^2:7=7
d)5^140:25^70+8^32:16^24+100^25:10^48-2^2x5^2=
=5^140:5^140+2^96:2^96+10^50:10^48-10^2=
=1+1+10^2-10^2=2
e)[3^100:3^10+(9^5x3^14)^5:27^10+(4^30:4^29-1^20)^102:81^3]: :(2^10:2^9+1)^90=
=[3^90+(3^10x3^14)^5:27^10+(4^30:4^29-1^20)^102:81^3]: :(2^10:2^9+1)^90=
=[3^90+(3^24)^5:3^30+(4^1-1)^102:3^12]:(2+1)^90=
=[3^90+3^120:3^30+3^102:3^12]:3^90=
=[3^90+3^90+3^90]:3^90=
=1+1+1=3







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