Matemáticas, pregunta formulada por soloquieroponerxd, hace 1 mes

es para ahora doy puntos y coronita , necesito procedimiento

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Contestado por peachiiipeach
1

a. 6^3 * 6 * 6^5 = 6^9\\\\b. (-7)^9 : ( - 7)^5 = (-7)^4 = 7^4\\\\c. ( ( - 2 )^4)^3 = ( - 2) ^1^2 = 2 ^1 ^2\\\\d. (-2)^9 : ((-2)^4 * ( - 2)^9)=(-2)^9 : ( - 2) ^1^3 = 2^-^4\\\\e. 20^5 : ( - 4) ^5 = -5^5\\\\f. (-36)^4 : ( -9)^4 = 4^4\\\\g. 12^9 : ((-3)^9 * (-4)^9) = 12^9 : 12^9 = 1\\\\h. ( - 5)^0 * ( 20^8 : ( - 4)^8) = 1 * ( - 5)^8 = 5^8\\\\i. (6^3)^2 * (( -7)^5 * ( - 7)) = 6^6 * ( - 7) ^6 = ( - 42) ^6 = 42^6\\\\j. (5^7)^4 * (5^4)^3 = 5^2^8 * 5^1^2 = 5^4^0\\\\

k. (-5)^8 : ( -5)^4 * ( - 5) = (-5)^4 * (-5) = -5^5\\\\l. (8^5)^4 : ((-2)^1^2 * (-4) ^1^2 = 8^2^0 : 8^1^2 = 8^8\\\\m. (a^5)^3 : (a^7)^2 = a^1^5 : a^1^4 = a\\\\n. 2^5 * 8 * 2^7 * 16 = 2^5 * 2^3 * 2^7 * 2^4 = 2^1^9\\\\o. (3^2)^3 * 27 * 9^4 = 3^6 * 3^3 * (3^2) ^4 = 3^9*3^8=3^1^7\\\\p. 3^1^5 : 81^3 = 3^1^4 : (3^4)^3 = 3^1^5 : 3^1^2 = 3^3\\\\q. (m^3 * m) ^2 : (m^2)^4 = (m^4)^2 : m^8 = m^8 : m^8 = 1\\\\r. (-10)^2^0 : ((-2)^9 * 5^9)^2 =10^2^0 : (-10^9 )^2 = 10^2^0 : 10^1^8 = 10^2\\\\

s. ((-3)^5 * (-2)^5)^3 * (6^4)^2 = ( 6^5)^3 * 6^8 = 6^1^5 * 6^8 = 6^2^3\\\\t. (a^3 * a^4)^2 : (a^1^5 : a^1^3) ^2 = (a^7) ^2 : (a^2) ^2 = a^1^4 : a^4 = a^1^0\\\\u. ((-30)^1^9 : 5^1^9) : ((-2)^4 * 3^4)^3 = -6^1^9 : ( -6^4)^3 = -6^1^9 : ( - 6)^1^2 = -6^7\\\\v. m^1^4 : (m^8 * m) = m^1^4 : m^9 = m^5

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