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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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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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