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

Toán Lớp 9: a) (2x/ căn (5) – căn (3) ) – (2x/ căn(3)+ 1)= căn (5) +1 b) x – 6căn (x-3) -10=0

Toán Lớp 9: a) (2x/ căn (5) – căn (3) ) – (2x/ căn(3)+ 1)= căn (5) +1
b) x – 6căn (x-3) -10=0

Comments ( 2 )

  1. Đáp án:
     
    Giải thích các bước giải:
     

    toan-lop-9-a-2-can-5-can-3-2-can-3-1-can-5-1-b-6can-3-10-0

  2. Giải đáp+Lời giải và giải thích chi tiết:
     a) \frac{2x}{\sqrt{5} – \sqrt{3}} – \frac{2x}{\sqrt{3} + 1} = \sqrt{5} + 1
    ⇔ \frac{2x(\sqrt{5} + \sqrt{3})}{(\sqrt{5} + \sqrt{3})(\sqrt{5} – \sqrt{3})} – \frac{2x(\sqrt{3} – 1)}{(\sqrt{3} + 1)(\sqrt{3} – 1)} = \sqrt{5} + 1
    ⇔ \frac{2\sqrt{5}x + 2\sqrt{3}x}{2} – \frac{2\sqrt{3}x – 2x}{2} = \frac{2(\sqrt{5} + 1}{2}
    ⇒ 2\sqrt{5} + 2\sqrt{3}x – 2\sqrt{3}x + 2x = 2\sqrt{5} + 2
    ⇔ 2\sqrt{5}x + 2x – 2\sqrt{5} – 2 = 0
    ⇔ 2\sqrt{5}(x – 1) + 2(x – 1) = 0
    ⇔ (2\sqrt{5} + 2)(x – 1) = 0
    ⇒ x – 1 = 0
    ⇔ x = 1
    b) x – 6\sqrt{x – 3} + 10 = 0
    ⇔ x – 3 – 6\sqrt{x – 3} + 9 + 4 = 0
    ⇔ (\sqrt{x – 3} – 3)² + 4 = 0
    ⇔ (\sqrt{x – 3} – 3)² = -4    [Vô nghiệm vì (\sqrt{x – 3} – 3)² ≥ 0]

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