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

Toán Lớp 8: cho 5 số dương a1+a2+a3+a4+a5=1.cmr:(1/a1-1)(1/a2-1)1/a3-1)(1/a4-1)(1/a5-1)>=1024

Toán Lớp 8: cho 5 số dương a1+a2+a3+a4+a5=1.cmr:(1/a1-1)(1/a2-1)1/a3-1)(1/a4-1)(1/a5-1)>=1024

Comments ( 1 )

  1. Giải đáp:
     
    Lời giải và giải thích chi tiết:
    $\dfrac{1}{a_{1}} – 1 = \dfrac{1 – a_{1}}{a_{1}} $
    $ = \dfrac{a_{2} + a_{3} + a_{4} + a_{5}}{a_{1}} ≥ \dfrac{4\sqrt[4]{a_{2}a_{3}a_{4}a_{5}}}{a_{1}} (1)$
    $\dfrac{1}{a_{2}} – 1 = \dfrac{1 – a_{2}}{a_{2}} $
    $ = \dfrac{a_{1} + a_{3} + a_{4} + a_{5}}{a_{2}} ≥ \dfrac{4\sqrt[4]{a_{1}a_{3}a_{4}a_{5}}}{a_{2}} (2)$
    $\dfrac{1}{a_{3}} – 1 = \dfrac{1 – a_{3}}{a_{3}} $
    $ = \dfrac{a_{1} + a_{2} + a_{4} + a_{5}}{a_{1}} ≥ \dfrac{4\sqrt[4]{a_{1}a_{2}a_{4}a_{5}}}{a_{3}} (3)$
    $\dfrac{1}{a_{4}} – 1 = \dfrac{1 – a_{4}}{a_{4}} $
    $ = \dfrac{a_{1} + a_{2} + a_{3} + a_{5}}{a_{4}} ≥ \dfrac{4\sqrt[4]{a_{1}a_{2}a_{3}a_{5}}}{a_{4}} (4)$
    $\dfrac{1}{a_{5}} – 1 = \dfrac{1 – a_{5}}{a_{5}} $
    $ = \dfrac{a_{1} + a_{2} + a_{3} + a_{4}}{a_{5}} ≥ \dfrac{4\sqrt[4]{a_{1}a_{2}a_{3}a_{4}}}{a_{5}} (5)$
    $(1).(2).(3).(4).(5):$
    $ (\dfrac{1}{a_{1}} – 1)(\dfrac{1}{a_{2}} – 1)(\dfrac{1}{a_{3}} – 1)(\dfrac{1}{a_{4}} – 1)(\dfrac{1}{a_{1}} – 1) ≥ 4^{5}.\dfrac{a_{1}a_{2}a_{3}a_{4}a_{5}}{a_{1}a_{2}a_{3}a_{4}a_{5}} = 1024$

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

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