Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Simplify by taking roots of the numerator and denominator. Assume all variables represent positive numbers, which means no absolute value brackets. So this one we have an index of understood too, because there was no number with the square root. So all we're going to do is we're going to think about what's sqrt 100 and sqrt 81. Sqrt 100 is 10 and sqrt 81 is 9 because 10 * 10 is 109 * 9 is 81343. That one might be a little harder to see. Let's see, 343 is divisible by 7/7 into 34 goes four times, that's 28. And then that leaves us 63, so 49, and that's seven and seven. So the cube root of 7 ^3 / 1000 could be thought of as 10 ^3. So the index of three goes into the exponent of three one time. So we got a final answer of 7/10 sqrt 25 is going to be 5. Two goes into the exponent of five twice with one leftover, and the index of two goes into the exponent of 6/3 times. So that's our final answer. This next 181 we need to prime factor 9 * 9, three, three, and three and three. So there are four threes. So the 4th root of 3 to the 4th, X to the 4th over, Y to the 8th, Z to the 4th. We just put the index into the exponents, so we get 3X over y ^2 C as a final answer. This last 1-6 goes into 91 time with three leftover. Six goes into 12 twice with. None leftover. Six goes into 13 twice with one leftover. So we could actually make that a great big radical. Here's a few for you to try. The next one we're going to divide and if possible, we're going to simplify. So the first thing I would do is I would always recommend to simplify the inside once we have it written as a division so we can see the wise cancel. And seven goes into 28 four times and sqrt 4 is just two. Here once again, seven goes into 56. Eight times the A's are going to cancel and we get b ^3. Well, eight we could think of as 2 * 4 or 2 * 2 * 2. So we're going to have sqrt 2 ^3 b ^3. The index of two goes into the exponent of three one time with the one leftover. Index of two goes into exponent 3 one time with one leftover. So that's our answer. So seven goes into 189, seven goes into 18 twice, and then seven goes into 49 seven times. X to the fifth over X ^2. It's going to leave us an X ^3. Y to the 7th over y ^2 is going to give us AY to the fifth. Remember, index is still 3, so the cube root of 27 is 3, the cube root of X ^3 is X, and then three goes into the exponent of five, one full time with two leftover. This last 164 / 2 is going to give us 32A to the 11th over A is going to give us A to the 10th and B to the 28th over B to the -2. Remember, when we're dividing, we really subtract. So 28 -. A negative two would give me 30. The index again was 5 S 32 we could think of as 4 * 8 or two times 242. There's going to be 5 twos there, so the 5th root of 2 to the 5th, A to the 10th, B to the 30th. Literally the five's just going to go into each of the exponents. So we get two a ^2 b to the 6th. Looking at the next page, we've got a few for you to try, and then we're going to rationalize the denominator. When we say rationalize the denominator, we need to have the square root outside. We're not within the denominator or cube roots. So I don't know sqrt 7, but I do know sqrt 49. So I'm going to multiply the top and the bottom each by sqrt 7, and I'm going to get sqrt 42 / sqrt 49 and sqrt 49 we know is 7, so we're going to get sqrt 42 / 7 as our final answer. This next one, we're going to multiply the top and bottom just by sqrt 2. So we're going to get 3 square roots of 10 / 2 square roots of four. But sqrt 4 is 2, SO 3 sqrt 10 / 2 * 2, or three sqrt 10 / 4. This next one gets tricky. I've actually got to have three of everything because of the index of three. I currently only have one seven, so I'm going to multiply by two more sevens. I need to have a multiple of three in the exponent. I currently have two BS b ^2, so I'm going to multiply by one more B. And if I do it in the bottom, I have to do it in the top. So what happens here? In the denominator, we get cube root of 7, cube b ^3 and the index goes into the exponents evenly. So we're just going to get 7B on the top. We're going to get the cube root of 3 * 7 ^2 * A to the 4th times B. So the three doesn't go into the exponent of on the three one evenly, and the three doesn't go into two evenly. The three goes into four though, one time with one leftover. So we're going to have 3 * 7 ^2 * a * b and 3 * 7 ^2 3 times sevens. 2121 * 7 is 147, so a cube root of 147 AB over 7B would be our final. This next one it's the same concept. The index is 3, so I need three of I need multiples of three in the bottom. So I'm going to have an A and AB squared. And if I multiply the bottom by an A and AB squared and this is still the cube root. If we do the top, we got to do the bottom, so we get cube root of 5 AB squared on the top over a cube b ^3. So our final answer would be the cube root of 5 AB squared over AB. The next 132. If we thought about 3232 four times, 822 eights, 4:00 and 2:00. So we get 5 twos here. Well I need to have 6 twos because I need the index of two to evenly go in. So if I currently have two to the 5th a ^2 BI need to have a multiple of the index. So I'm going to multiply by a, two and AB because two goes into the exponent on the a evenly, but two doesn't go into the exponent on the B evenly. And if I multiply the bottom by two B, I'm going to multiply the top by 2B. So we're going to get sqrt 14 B over sqrt 2 to the six a ^2 b ^2. So that's going to give us a final answer. Sqrt 14 B over two goes into the exponent of 6/3 times. And two cubed is going to be 8 AB. This next one, 21 and 75, those are both divisible by three. So I might start by reducing. So 21 and 75 divisible by three, three 7 / 25. And X ^2 / X leaves me an X on the top. And AY over Y to the 4th leaves me y / y to the fifth leaves me Y to the 4th. Well, this one actually, I don't have to even rationalize. I just have to simplify because we know sqrt 25 is five. And the square root of Y to the 4th is y ^2. So our final here is going to be sqrt 7 X over five y ^2. Got a few for you to try. Then we're going to switch and we're going to rationalize the numerators instead of the denominators. So if I want to rationalize the numerator, now I've got to get square root out of the top. So I'm going to multiply by root 10 over root 10, and I'm going to get sqrt 10 ^2 / sqrt 30 X or 10 / sqrt 30 X This next one, it's cube root, so I need to have three of them. So I only have one five right now. So I'm going to multiply by 5 ^2. If I do the top, I've got to do the bottom, so that makes the top turn into the cube root of 5 ^3 over the cube root of 25 * 4 a hundred. So our final answer here would be 5 cube roots 5 over cube roots of 100. Thank you and have a wonderful day.