All Answers 358095 Answers

We have a couple of things to consider

First.....the domain of the function ih the numerator is  x ≥ 0

For the  function in the denominator.....1 - x^2  must be greater than 0

So

1 - x^2 > 0

( 1 - x) ( 1 + x ) >  0

If     -1 < x < 1     this is true

But....the restriction in the denominator makes the true domain   0  ≤ x < 1

√ [3x] - 4√3               2√[2x ]  + √2

__________   =       ____________

√x  - √2                      √[6x] - 2√3

√ [3x] - √48              √[8x ]  + √2

__________   =       ____________     cross multiply

√x  - √2                      √[6x] - √12

( √ [3x] - √48 ) (  √[6x] - √12) =  (√x  - √2) ( √[8x ]  + √2)     simplify

3x√2  - 12√[2x] - 6√x + 24   =  2x√2 -  4√x + √[2x] - 2

x√2 - 13√[2x]  - 2√x + 26  = 0

√2 √x √x -  13√2 √x  - 2√x + 26 = 0      factor as

√2√x [ √x  - 13 ] - 2 [ √x - 13]  = 0

[ √x - 13]  [ √[2x] - 2 ]  = 0

So either

√x - 13  = 0                                   or         √[2x] - 2  = 0

√x = 13   square both sides                       √[2x]  = 2        square both sides

x = 169                                                          2x   = 4

                                                                       x = 2

Reject the second solution....it makes the original denominators = 0

So

x = 169

I'll get you started :)

\(\dfrac{\sqrt{3x}-4\sqrt{3}}{\sqrt{x}-\sqrt{2}} =\dfrac{2\sqrt{2x}+\sqrt{2}}{\sqrt{6x}-2\sqrt{3}}\\ \dfrac{\sqrt{x}+\sqrt{2}}{\sqrt{x}+\sqrt{2}}\cdot\dfrac{\sqrt{3x}-4\sqrt{3}}{\sqrt{x}-\sqrt{2}} =\dfrac{2\sqrt{2x}+\sqrt{2}}{\sqrt{6x}-2\sqrt{3}}\cdot \dfrac{\sqrt{6x}+2\sqrt{3}}{\sqrt{6x}+2\sqrt{3}}\)

When you expand the each side out you will have no irational terms on the bottom so it will then be much easier.

If you get stuck you can let us know (with an explanation)

Solve for x:
3 (sqrt(x) - 4) (sqrt(2) sqrt(x) - 2) = -sqrt(2) (sqrt(2) - sqrt(x)) (2 sqrt(x) + 1)

Add sqrt(2) (2 sqrt(x) + 1) (sqrt(2) - sqrt(x)) to both sides:

sqrt(2) (sqrt(2) - sqrt(x)) (2 sqrt(x) + 1) + 3 (sqrt(x) - 4) (sqrt(2) sqrt(x) - 2) = 0

sqrt(2) (sqrt(2) - sqrt(x)) (2 sqrt(x) + 1) + 3 (sqrt(x) - 4) (sqrt(2) sqrt(x) - 2) = 26 + (-13 sqrt(2) - 2) sqrt(x) + sqrt(2) x:
26 + (-13 sqrt(2) - 2) sqrt(x) + sqrt(2) x = 0

Simplify and substitute y = sqrt(x).
26 + (-13 sqrt(2) - 2) sqrt(x) + sqrt(2) x = 26 + (-13 sqrt(2) - 2) sqrt(x) + sqrt(2) (sqrt(x))^2
 = sqrt(2) y^2 + (-2 - 13 sqrt(2)) y + 26:
sqrt(2) y^2 + (-2 - 13 sqrt(2)) y + 26 = 0

The left hand side factors into a product with two terms:
(y - 13) (sqrt(2) y - 2) = 0

Split into two equations:
y - 13 = 0 or sqrt(2) y - 2 = 0

Add 13 to both sides:

y = 13 or sqrt(2) y - 2 = 0

Substitute back for y = sqrt(x):
sqrt(x) = 13 or sqrt(2) y - 2 = 0

Raise both sides to the power of two:
x = 169 or sqrt(2) y - 2 = 0

Add 2 to both sides:
x = 169 or sqrt(2) y = 2

Divide both sides by sqrt(2):
x = 169 or y = sqrt(2)

Substitute back for y = sqrt(x):
x = 169 or sqrt(x) = sqrt(2)

Raise both sides to the power of two:

x = 169               x = 2 is extraneous

We have a right triangle with   a hypotenuse of   6371 + 6.6  =  6377.6 km

The Earth's radius is one of the legs = 6371 km

And the distance to the horizon is the other leg and is given by :

√ [ 6377.6^2 -6371^2 ]  ≈   290.1 km

³√x^8                       x^(8/3)

__________   =    ________    =  x^(8/3) - x^(2/3) )   =  x^(6/3)   = x^2

 ( X^2)^(1/3)           x^(2/3)

I don't think 120 ways is right...

Sorry to be a nuisance but what would be the formula be for this please. In year 1 you can see there is a one off figure of 714 which changes year 1 total.

"a" can be any integer  because 2a  is always divisible by 2....the requirement for "eveness"

(f - y)  = [ 2x] - [ 3 - x ] =   3x - 3

(f - y) (4)  =   3(4) - 3   =  12 - 3   = 9

Yes I agree

I'd just write it as

\(f(x)=x(x+22)^3\)

(x+22)^3 (x) = 0      (?)

\(3! \sqrt{49} = 42\)