Kuis
Slmt thn baru (iiI)
αβγ = 24
α(β+γ) + βγ = 26
α + β + γ = 9
α < β < γ
γ² - β² - α² = ....
(A.) = α
(B.) = β
(C.) = γ
(D.) = 1
(E.) = 0
Slmt thn baru (iiI)
αβγ = 24
α(β+γ) + βγ = 26
α + β + γ = 9
α < β < γ
γ² - β² - α² = ....
(A.) = α
(B.) = β
(C.) = γ
(D.) = 1
(E.) = 0
Jawaban:
B. = β
Penjelasan dengan langkah-langkah:
αβγ = 24
α(β+γ) + βγ = 26
α + β + γ = 9
α < β < γ
γ² - β² - α² = ....
karena a<B<y maka bisa diperoleh 2<3<4 , kita subtitusikan untuk pembuktian
aBy = 24
(2)(3)(4) = 24 → terbukti
a(B+y) + By = 26
2(3+4) +(3.4)
14 + 12 = 26 → terbukti
a+B+y = 9
2+3+4 = 9 → terbukti
maka
4²-3²-2²
= 16-9-4
= 3 → β
B = β
#Penalaran
Aljabar
αβγ = 24
2 × 3 × 4 = 24 ✔
α + β + γ = 9
2 + 3 + 4 = 9 ✔
α(β + γ) + βγ = 26
2(3 + 4) + 3 × 4 = 26 ✔
α < β < γ
2 < 3 < 4
γ² - β² - α²
= 4² - 3² - 2²
= 16 - 9 - 4
= 3
γ² - β² - α² = β