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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 9: căn bậc 3 ( 9 + 4 căn 5 ) + căn bậc 3 ( 9 – 4 căn 5 )

Toán Lớp 9: căn bậc 3 ( 9 + 4 căn 5 ) + căn bậc 3 ( 9 – 4 căn 5 )

Comments ( 1 )

  1. Đặt: a = $\sqrt[3]{9 + 4\sqrt{5}}$    ⇒ a³ = 9 + 4\sqrt{5}
    b = $\sqrt[3]{9 – 4\sqrt{5}}$             ⇒ b³ = 9 – 4\sqrt{5}
    ⇒ a³ + b³ = 9 + 4\sqrt{5} + (9 – 4\sqrt{5}) = 18
         ab = ($\sqrt[3]{9 + 4\sqrt{5}}$)($\sqrt[3]{9 – 4\sqrt{5}}$)
              = $\sqrt[3]{(9 + 4\sqrt{5})(9 – 4\sqrt{5})}$
              = 1
    Ta có:
    (a + b)³ = a³ + b³ + 3a²b + 3ab²
                  = a³ + b³ + 3ab(a + b)
                  = 18 + 3.1.(a + b)
                   = 18 + 3(a + b)
    ⇒ (a + b)³ – 3(a + b) – 18 = 0
    ⇔ (a + b)³ – 3(a + b)² + 3(a + b)² – 9(a + b) + 6(a + b) – 18 = 0
    ⇔ (a + b)²(a + b – 3) + 3(a + b)(a + b – 3) + 6(a + b – 3) = 0
    ⇔ (a + b – 3)[(a + b)² + 3(a + b) + 6] = 0
    Vì: (a + b)² + 3(a + b) + 6 = [(a + b)² + 3/2 . 2 . (a + b) + 9/4] + 15/4 = (a + b + 3/2)² + 15/4 > 0
    ⇒ a + b – 3 = 0
    ⇔ a + b = 3
    ⇒ $\sqrt[3]{9 + 4\sqrt{5}}$ + $\sqrt[3]{9 – 4\sqrt{5}}$ = 3
     
     

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222-9+11+12:2*14+14 = ? ( )