3 Challenging Pre-Cal Questions - Need Help?

Question

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***Click the link the problems are there***

Answer 1

1.rationalise the Dr
(3rt2-6)(4rt2+3rt6)/(4rt2-3rt6)(4rt2+3rt6)
=3rt2(4rt2+3rt6)-6(4rt2+3rt6)/(4rt2)^2-(3rt6)^2
=24+9rt12-24rt2-18rt6/(32-54)
=24+18rt2-24rt2-18rt6/(-22)

2.2x-3>-4x/(x-1)
(2x-3)(x-1)>-4x
2x^2-5x+3>-4x
adding 4x
2x^2-x+3>0
2x^2+2x-3x+3>0
2x(x+1)-3(x+1)>0
(x+1)(2x-3)>0

3.f(f(x)=[3-2(3-2x/2-3x)]/[2-3(3-2x/2-3x)]
=3(2-3x)-2(3-2x)/2(2-3x)-9+6x)
=6-9x-6+4x/4-6x-9+6x
=-5x/-5
=x
hence the question

Answer 2

(3sqrt(2) - 6)/(4sqrt(2) - 3sqrt(6))

Same as

(sqrt(18) - sqrt(36))/(sqrt(32) - sqrt(54))

Multiply top and bottom by sqrt(32) + sqrt(54)

((sqrt(18) - sqrt(36))(sqrt(32) + sqrt(54))/((sqrt(32) - sqrt(54))(sqrt(32) + sqrt(54))

(sqrt(576) + sqrt(972) - sqrt(1152) - sqrt(1944))/(32 + sqrt(1728) - sqrt(1728) + 54)

(24 + sqrt(3 * 324) - sqrt(576 * 2) - sqrt(6 * 324))/(32 + 54)

(24 + 18sqrt(3) - 24sqrt(2) - 18sqrt(6))/86

(24(1 - sqrt(2)) + 18(sqrt(2) - sqrt(6)))/86

ANS :
(12(1 - sqrt(2)) + 9(sqrt(2) - sqrt(6)))/43
or
(12 - 12sqrt(2) + 9sqrt(2) - 9sqrt(6))/43

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2x - 3 > (-4x)/(x - 1)
I went to www.quickmath.com to solve.

I know that 1 would make this problem undefined and any answers of x = -n, would make this problem not true

x > 1

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f(x) = (3 - 2x)/(2 - 3x)

f(f(x)) = f((3 - 2x)/(2 - 3x))

f((3 - 2x)/(2 - 3x)) = (3 - 2((3 - 2x)/(2 - 3x)))/(2 - 3((3 - 2x)/(2 - 3x)))

f(f(x)) = (3 - ((6 - 4x)/(2 - 3x)))/(2 - ((9 - 6x)/(2 - 3x)))

f(f(x)) = ((3(2 - 3x) - (6 - 4x))/(2 - 3x)) / ((2(2 - 3x) - (9 - 6x))/(2 - 3x))

f(f(x)) = ((6 - 9x - 6 + 4x)/(2 - 3x))/((4 - 6x - 9 + 6x)/(2 - 3x))
f(f(x)) = (-5x/(2 - 3x))/(-5/(2 - 3x))
f(f(x)) = (-5x/(2 - 3x))*((2 - 3x)/(-5))
f(f(x)) = (-5x(2 - 3x))/(-5(2 - 3x))
f(f(x)) = x

Answer 3

[3sqrt(2) -6]/[4sqrt(2) -3sqrt(6)]
Multiply both numerator and denominator by 4sqrt(2) +3sqrt(6)
This gives [12-3sqrt(2) -9sqrt(6)]/-11

2x-3 > -4x/(x-1)
(2x-3)(x-1) < -4x
2x^2-2x -3x +3 +4x <0
2x^2-x+3 < 0
There are no real values of x that satisfy this inequality.

f(x)= (3-2x)/(2-3x)
f(f(x)= [3-2(3-2x)/(2-3x)]/ [2-3(3-2x)/2-3x)] =
[3 - (6+4x)/(2-3x)]/[2 - (9+6x)/(2-3x)] =
[6-9x-6 +4x]/(2-3x)/[4-6x -9 +6x]/(2-3x) =
(-5x)/(2-3x) * (2-3x)/(-5) = (-5x)/-5 = x

Answer 4

3rt2 -6 = 3(rt2-2) = 3rt2(1-rt2)
4rt2 -3rt6 = rt2(4 - 3rt3)
so num/denom = 3(1-rt2)/(4-3rt3)

Answer 5

pl. state your question

Answer 6

hard