simplyfiy each expression. leave in radicle form. use absolute value symbols if neccissary?

Question

1.√8ab^4
2. 5√3÷9√3
3. 14√5-7√5
4. 2√2+2√8
5. √24+√54
6. √63-√112+√121

Answer 1

1. √(8a)b
2. 5/9
3. 7√(5)
4. 2√(2)+2√(8)
5. 5√(6)
6. - 7√(7)+11

hope this helps

Answer 2

1. √(8ab^4) = 2b²√(2a)
2. 5√3 ÷ 9√3 = 5/9
3. 14√5 - 7√5 = 7√5
4. 2√2 + 2√8 = 2√2 + 4√2 = 6√2
5. √24 + √54 = 2√6 + 3√6 = 5√6
6. √63 - √112 + √121 = 3√7 - 4√7 + 11 = 11 - √7

Answer 3

???????????????????? ordinary roots ... cube root, 5th root, etc may well be adverse. as an occasion, ? ?-?8? ? = -2, because of the fact (-2)³ = -8. So, for all values of x, ? ?x?³? ? = x although: Even roots ... sq. root, 4th root, 6th root, etc., won't be able to be adverse. as an occasion, ? ?1?6? ? ? = 4 that's helpful. ?????????? ?????????? ?????????? ?? ? ?1?6? ? ? ?? -?4?? because of the fact sq. roots are in no way adverse. So, ? ?x?²? ? ? = ?x? that's non adverse. ?????????? ? ?x?²? ? ? ?? x?? because of the fact x may well be adverse. i wish that's no longer puzzling. ???????????????????? ? ?????? ? ?4?0?0?x?²?y??? ? ? = ? ?2?0?²?x?²?(?y?³?)?²? ? ?????????? ? sq. roots are in no way adverse !! ? 20xy³?????? if x and y may well be adverse, then x and y³ can the two be adverse. ?????????? ?????????? ?????????? ????? So, this is faulty. Absolute fee is needed to cause them to non adverse. = 20?xy³? ?????????? ? answer ?? (b.) ???????????????????? ? ?????? ? ?-?1?2?5?a??? ? ? ?????????? ? cube roots may well be adverse. So, absolute fee isn't needed. = ? ?(?-?5?)?³?(?a?³?)?³? ? ? = -5a³ ?????????? ? answer ?? (c.) ???????????????????? ? ?????? ? ?8?1?x???y???? ? ??????????? even root ????? won't be able to be adverse ????? might desire to apply absolute fee = ? ?3???x???x??(??y?²?)???y? ? ? 3xy²? ??x???y? ? ?????????? ? if x and y may well be adverse, ?????????? ?????????? ?????????? ?????????? ???????? then x may well be adverse so as that absolute fee is needed for x. ?????????? ?????????? ?????????? ?????????? ???????? yet, y² is in no way adverse. So, absolute fee isn't needed for y². = 3?x?y²? ??x???y? ?????????? ? answer ?? (a.) ???????????????????? ? ?????? ? ?6?4?a???b?²? ? ? ?????????? ? ordinary roots may well be adverse. So, absolute fee isn't needed. = ? ?4?³?(?a?²?)??³?b?²? ? ? = 4a²? ??b?²? ? ? ?????????? ? answer ?? (d.) ????????????????????