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Maths problem please help!!!?

1.Evaluate under root a + 1/ underoot a

if a=7-4 under root three..

2.2^x-1+2^x+1=320

please help to solve these problems..i'll be thnkful

regards

1 Answer

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  • ?
    Lv 7
    8 years ago
    Favorite Answer

    1.

    a = 7 - 4 sqrt(3)

    Express a as a square:

    a = 4 - 4 sqrt(3) + (sqrt(3))^2

    a = (2 - sqrt(3))^2

    Therefore, a = 2 - sqrt(3)

    So sqrt(a) + 1 / sqrt(a) can be written as:

    (2 - sqrt(3)) + 1 / (2 - sqrt(3))

    Rationalize the denominator of the second term by multiplying by (2 + sqrt(3)) / (2 + sqrt(3)).

    (2 - sqrt(3)) + (2 + sqrt(3)) / ((2 - sqrt(3))(2 + sqrt(3))) =

    (2 - sqrt(3)) + (2 + sqrt(3)) / (2^2 - (sqrt(3))^2) =

    (2 - sqrt(3)) + (2 + sqrt(3)) / (4 - 3) =

    (2 - sqrt(3)) + (2 + sqrt(3)) / 1 =

    (2 - sqrt(3)) + (2 + sqrt(3)) =

    2 + 2 - sqrt(3) + sqrt(3) =

    2 + 2 =

    4

    Answer: 4

    2.

    2^(x-1) + 2^(x+1) = 320

    2^(x-1) + 2^((x - 1) + 2) = 320

    2^(x - 1) + 2^(x - 1) * 2^2 = 320

    2^(x - 1) + 2^(x - 1) * 4 = 320

    2^(x - 1) + 4 * 2^(x - 1) = 320

    5 * 2^(x - 1) = 320

    2^(x - 1) = 320/5

    2^(x - 1) = 64

    2^(x - 1) = 2^6

    x - 1 = 6

    x = 7

    Answer: 7

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