√(54 + 14√5) + √(12 - 2√35) + √(32 - 10√7)
Matematika
fitrikurniarahmawati
Pertanyaan
√(54 + 14√5) + √(12 - 2√35) + √(32 - 10√7)
2 Jawaban
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1. Jawaban ErikCatosLawijaya
Materi : Akar dan pangkat
Rumus dasar :
√[a + b + 2√ab] = √a + √b
√[a + b - 2√ab] = √a - √b
Pembahasan :
√(54 + 14√5) + √(12 - 2√35) + √(32 - 10√7)
= √[54 + 2√(49.5)] + √[7 + 5 - 2√(7.5)] + √[32 - 2√(25.7)]
= √[54 + 2√245] + [√7 - √5] + √[32 - 2√175]
= √[49 + 5 + 2√(49.5)] + [√7 - √5] + √[25 + 7 - 2√(25.7)]
= [√49 + √5] + [√7 - √5] + [√25 - √7]
= 7 + √5 + √7 - √5 + 5 - √7
= 12 -
2. Jawaban Anonyme
Radicals.
If a ≥ 0, b ≥ 0, and c ≥ 0, then:
√(a + b + 2√ab) = √a + √b
√(a + b - 2√ab) = √a - √b with a > b
√(54 + 14√5) + √(12 - 2√35) + √(32 - 10√7)
= √(54 + 2√245) + √(12 - 2√35) + √(32 - 2√175)
= √[5 + 49 + 2√(5 × 49)] + √[7 + 5 - 2√(7 × 5)] + √[25 + 7 - 2√(25 × 7)]
= √5 + √49 + √7 - √5 + √25 - √7
= 7 + 5
= 12