(x^a2/x^b2)^1/a+b × (x^b2/x^c2)^1/b+c × (x^c2/x^a2)^1/c+a=1​

Question

(x^a2/x^b2)^1/a+b × (x^b2/x^c2)^1/b+c × (x^c2/x^a2)^1/c+a=1​

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Allison 3 weeks 2021-10-01T14:57:00+00:00 1 Answer 0 views 0

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    2021-10-01T14:58:25+00:00

    Answer:

    Prove that

    (

    x

    b

    2

    x

    a

    2

    )

    a+b

    1

    ×(

    x

    c

    2

    x

    b

    2

    )

    b+c

    1

    ×(

    x

    a

    2

    x

    c

    2

    )

    c+a

    1

    =1

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    Leta=(

    x

    b

    2

    x

    a

    2

    )

    a+b

    1

    ×(

    x

    c

    2

    x

    b

    2

    )

    c+b

    1

    ×(

    x

    a

    2

    x

    c

    2

    )

    a+c

    1

    =(x

    a

    2

    −b

    2

    )

    a+b

    1

    ×(x

    b

    2

    −c

    2

    )

    c+b

    1

    ×(x

    c

    2

    −a

    2

    )

    a+c

    1

    =x

    a+b

    (a+b)(a−b)

    ×x

    c+b

    (c+b)(b−c)

    ×x

    a+c

    (a+c)(c−a)

    =x

    a−b

    ×x

    b−c

    ×x

    c−a

    =x

    a−b+b−c+c−a

    =x

    0

    a=1

    ∴(

    x

    b

    2

    x

    a

    2

    )

    a+b

    1

    ×(

    x

    c

    2

    x

    b

    2

    )

    c+b

    1

    ×(

    x

    a

    2

    x

    c

    2

    )

    a+c

    1

    =1

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