f) (4π+π²) / 3π = [π(4+π)] / 3π = (4+π)/3 = [2(4+π)] / (2*3) = (8+2π)/6
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Bonsoir,Pour démontrer une égalité, souvent il faut calculer/développer le premier calcul pour arriver au 2ème.
a) (V5-2)² - 4 = (V5-2)² - 2² = (V5-2-2)(V5-2+2) = V5*(V5-4)
b) (6 + 9V2) / (5V2) = [(6 + 9V2)*V2] / [5V2*V2] = (6V2+18) / 10 = [2(3V2+9)] / (2*5) = (3V2+9) / 5
c) (V2 - 3)² = (V2)² - 6V2 + 3² = 2 - 6V2 + 9 = 11-6V2
d) 0,00016 / (2*10^-3)² = (16*10^-5) / (4*10^-6) = 4*10 = 40
e) 1 / (3+V7) = [1*(3-V7)] / [(3-V7)(3+V7)] = (3-V7) / [3²-(V7)²] = (3-V7) / (9-7) = (3-V7) / 2
f) (4π+π²) / 3π = [π(4+π)] / 3π = (4+π)/3 = [2(4+π)] / (2*3) = (8+2π)/6