5/(sq rt 3 - 1). Multiply the numerator and denominator by (sq rt 3 + 1)
5(sq rt 3 + 1) / (sq rt 3 - 1)(sq rt 3 + 1). (sq rt 3 - 1)(sq rt 3 + 1 = 3 - 1 = 2.
5(sq rt 3 + 1) / 2. Answer
Not sure which one you mean without parentheses but maybe it is as written. I'll do them all in case.... 5/Sqrt3 - 1 ... 5/((Sqrt3)-1) ..... 5/Sqrt(3-1)
5 / Sqrt3 - 1
5Sqrt3 / 3 - 1
Sqrt3(5/3) - 1
5/((Sqrt3) - 1)
5((Sqrt3) - 1) / ((Sqrt3)-1)((Sqrt3)-1)
(5Sqrt3 - 5) / (3 - Sqrt3 - Sqrt3 - 1)
(4Sqrt3 - 5) / (2 - 2Sqrt3)
(4Sqrt3 - 5) / 2(1 - Sqrt3)
(1 + Sqrt3)(4Sqrt3 - 5) / 2(1 - Sqrt3)(1 + Sqrt3)
(4Sqrt3 - 5 + 4(3) - 5Sqrt3) / (2 + 2Sqrt3)(1 - Sqrt3)
(4Sqrt3 - 5 + 12 - 5Sqrt3) / (2 + 2Sqrt3)(1 - Sqrt3)
(7 - Sqrt3) / (2 - 2Sqrt3 + 2Sqrt3 - 2Sqrt9)
(7 - Sqrt3) / (2 - 2(3))
(7 - Sqrt3) / (2 - 6)
(7 - Sqrt3) / -4
-(7 - Sqrt3) / 4
(Sqrt3 - 7) / 4
Most of this could have been avoided... by multiplying by (Sqrt3 + 1) instead of (Sqrt3 -1).
5 / Sqrt(3-1)
5(Sqrt(3 - 1) / Sqrt(3 - 1)Sqrt(3 - 1)
5Sqrt2 / Sqrt2(Sqrt2)
5Sqrt2 / Sqrt4
5Sqrt2 / 2
Sqrt2(5/2)
a million / sqrt(3) => (a million / sqrt(3)) * (sqrt(3) / sqrt(3)) => (a million * sqrt(3)) / (sqrt(3) * sqrt(3)) => sqrt(3) / 3 9 / (sqrt(14) - 2) => (9 / (sqrt(14) - 2)) * ((sqrt(14) + 2) / (sqrt(14) + 2)) => (9 * (sqrt(14) + 2)) / ((sqrt(14) - 2) * (sqrt(14) + 2)) => (9 * (2 + sqrt(14))) / (sqrt(14) * sqrt(14) - 2 * sqrt(14) + 2 * sqrt(14) - 2 * 2) => 9 * (2 + sqrt(14)) / (14 - 4) => 9 * (2 + sqrt(14)) / 10 sqrt(8) / 2 => sqrt(4 * 2) / 2 => sqrt(4) * sqrt(2) / 2 => 2 * sqrt(2) / 2 => sqrt(2) (sqrt(5) + sqrt(3)) / 3 The denominator is already rationalized and you will't combine sqrt(5) and sqrt(3) jointly. even nonetheless, we are able to rationalize the numerator ((sqrt(5) + sqrt(3)) * (sqrt(5) - sqrt(3))) / (3 * (sqrt(5) - sqrt(3))) => (5 - 3) / (3 * (sqrt(5) - sqrt(3))) => 2 / (3 * (sqrt(5) - sqrt(3)))
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Verified answer
5/(sq rt 3 - 1). Multiply the numerator and denominator by (sq rt 3 + 1)
5(sq rt 3 + 1) / (sq rt 3 - 1)(sq rt 3 + 1). (sq rt 3 - 1)(sq rt 3 + 1 = 3 - 1 = 2.
5(sq rt 3 + 1) / 2. Answer
Not sure which one you mean without parentheses but maybe it is as written. I'll do them all in case.... 5/Sqrt3 - 1 ... 5/((Sqrt3)-1) ..... 5/Sqrt(3-1)
5 / Sqrt3 - 1
5Sqrt3 / 3 - 1
Sqrt3(5/3) - 1
5/((Sqrt3) - 1)
5((Sqrt3) - 1) / ((Sqrt3)-1)((Sqrt3)-1)
(5Sqrt3 - 5) / (3 - Sqrt3 - Sqrt3 - 1)
(4Sqrt3 - 5) / (2 - 2Sqrt3)
(4Sqrt3 - 5) / 2(1 - Sqrt3)
(1 + Sqrt3)(4Sqrt3 - 5) / 2(1 - Sqrt3)(1 + Sqrt3)
(4Sqrt3 - 5 + 4(3) - 5Sqrt3) / (2 + 2Sqrt3)(1 - Sqrt3)
(4Sqrt3 - 5 + 12 - 5Sqrt3) / (2 + 2Sqrt3)(1 - Sqrt3)
(7 - Sqrt3) / (2 - 2Sqrt3 + 2Sqrt3 - 2Sqrt9)
(7 - Sqrt3) / (2 - 2(3))
(7 - Sqrt3) / (2 - 6)
(7 - Sqrt3) / -4
-(7 - Sqrt3) / 4
(Sqrt3 - 7) / 4
Most of this could have been avoided... by multiplying by (Sqrt3 + 1) instead of (Sqrt3 -1).
5 / Sqrt(3-1)
5(Sqrt(3 - 1) / Sqrt(3 - 1)Sqrt(3 - 1)
5Sqrt2 / Sqrt2(Sqrt2)
5Sqrt2 / Sqrt4
5Sqrt2 / 2
Sqrt2(5/2)
a million / sqrt(3) => (a million / sqrt(3)) * (sqrt(3) / sqrt(3)) => (a million * sqrt(3)) / (sqrt(3) * sqrt(3)) => sqrt(3) / 3 9 / (sqrt(14) - 2) => (9 / (sqrt(14) - 2)) * ((sqrt(14) + 2) / (sqrt(14) + 2)) => (9 * (sqrt(14) + 2)) / ((sqrt(14) - 2) * (sqrt(14) + 2)) => (9 * (2 + sqrt(14))) / (sqrt(14) * sqrt(14) - 2 * sqrt(14) + 2 * sqrt(14) - 2 * 2) => 9 * (2 + sqrt(14)) / (14 - 4) => 9 * (2 + sqrt(14)) / 10 sqrt(8) / 2 => sqrt(4 * 2) / 2 => sqrt(4) * sqrt(2) / 2 => 2 * sqrt(2) / 2 => sqrt(2) (sqrt(5) + sqrt(3)) / 3 The denominator is already rationalized and you will't combine sqrt(5) and sqrt(3) jointly. even nonetheless, we are able to rationalize the numerator ((sqrt(5) + sqrt(3)) * (sqrt(5) - sqrt(3))) / (3 * (sqrt(5) - sqrt(3))) => (5 - 3) / (3 * (sqrt(5) - sqrt(3))) => 2 / (3 * (sqrt(5) - sqrt(3)))