(1 over radical three) over three - I assume that's how you are describing it - can be simplified like any division problem involving fractions in the numerator and denominator. To solve the problem, you take the fraction in the denominator (in this case 3 over 1) and invert it (making it 1 over 3) and multiply it by the numerator. You would have (1 over radical 3) times (1 over 3) which would give you 1 over (3 times radical 3)
2b
12x^2√5x
2a+5
sqrt(27) = 3*sqrt(3).
15
To simplify the expression radical 6 minus 4 radical 6, we first combine like terms. Since both terms have the same radical part (radical 6), we can subtract the coefficients in front of the radicals. This gives us -3 radical 6 as the simplified answer.
If you multiplied all the terms by y2, you could simplify this to -3y3.
Radical 147 simplified is 7 radical 3. radical147= radical 49* radical 3 the square root of 49 is 7 therefore the answer is 7 radical 3
3^3*radical(128) = 3^3*radical(2^7) = 3^3*radical(2^6*2) =3^3*2^3*radical(2) = 216*radical(2).
(1 over radical three) over three - I assume that's how you are describing it - can be simplified like any division problem involving fractions in the numerator and denominator. To solve the problem, you take the fraction in the denominator (in this case 3 over 1) and invert it (making it 1 over 3) and multiply it by the numerator. You would have (1 over radical 3) times (1 over 3) which would give you 1 over (3 times radical 3)
2b
12x^2√5x
A Huge ASS
minus a minus is plus, so it's -8 + 3, which is 3 - 8, or -5
2a+5
sqrt(27) = 3*sqrt(3).