This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Ili kutatua swali hili, tutatumia njia ya mambo (factor method) kupata mambo makuu ya nambari ndani ya mzizi wa mraba, kisha tutarahisisha. Sehemu ya 1: Tafuta mambo makuu ya 5184 Tutagawanya 5184 kwa nambari mkuu ndogo zaidi hadi tupate 1. 5184 ÷ 2 = 2592 2592 ÷ 2 = 1296 1296 ÷ 2 = 648 648 ÷ 2 = 324 324 ÷ 2 = 162 162 ÷ 2 = 81 Sasa 81 haigawanyiki kwa 2. Tunajaribu 3. 81 ÷ 3 = 27 27 ÷ 3 = 9 9 ÷ 3 = 3 3 ÷ 3 = 1 Kwa hiyo, 5184 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 = 2^6 × 3^4. Sehemu ya 2: Tafuta mambo makuu ya 49 49 ÷ 7 = 7 7 ÷ 7 = 1 Kwa hiyo, 49 = 7 × 7 = 7^2. Sehemu ya 3: Badilisha mambo makuu kwenye usemi Usemi asili ni sqrt(5184 × 49). Kubadilisha mambo makuu tunapata: sqrt((2^6 × 3^4) × 7^2) Sehemu ya 4: Rahisisha mzizi wa mraba Tunatumia kanuni kwamba sqrt(a × b) = sqrt(a) × sqrt(b) na sqrt(x^n) = x^n/2. sqrt(2^6 × 3^4 × 7^2) = sqrt(2^6) × sqrt(3^4) × sqrt(7^2) = 2^6/2 × 3^4/2 × 7^2/2 = 2^3 × 3^2 × 7^1 Sehemu ya 5: Andika jibu katika mfumo wa nguvu Jibu tayari liko katika mfumo wa nguvu za mambo makuu. 2^3 × 3^2 × 7^1 That's 2 down. 3 left today — send the next one.