How would i complete these problems?
1. (√6mn)^5
2. ^3√16x - ^3√2x^4
3.^4√x • ^3√2x
4. ^3√72x^8
5. √63a^5b • √27a^6b^4
and are these problems correct?
6. √2025xy/3√3 = 15√xy/√3
7. 3√18 + 8√50 = 49√2
8. √12 • √135 = 18√5
2 answers
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6, 7 and 8 are correct.
easier to write sqrt as fractional power for these
like sqrt (x) = x^(1/2)
1.
{(6 n m )^.5}^5 = (6 m n)^(5/2)
= (6 m n)^4/2 * (6 m n)^1/2
= 36 m^2 n^2 sqrt(6 m n)
2.
(16x)^1/3) - (2x^4)^1/3 = 2 (2)^1/3 (x)^1/3 - 2^1/3 x^4/3
= 2^1/3 x^1/3( 2 - x)
3.
x^1/4 2^1/3 x^1/3
= 2^1/3 x^7/12
4.
72^1/3 x^8/3
(2^3 * 3^2)^1/3 x^8/3
2 (9)^1/3 x^8/3
5.
(3*3*7 *a^5 * b^1 *3^3 * a^6* b^4)^1/2
= (3^5 a^11 b^5)^1/2
= 9 a^5 b^2 (3 a b)^1/2
easier to write sqrt as fractional power for these
like sqrt (x) = x^(1/2)
1.
{(6 n m )^.5}^5 = (6 m n)^(5/2)
= (6 m n)^4/2 * (6 m n)^1/2
= 36 m^2 n^2 sqrt(6 m n)
2.
(16x)^1/3) - (2x^4)^1/3 = 2 (2)^1/3 (x)^1/3 - 2^1/3 x^4/3
= 2^1/3 x^1/3( 2 - x)
3.
x^1/4 2^1/3 x^1/3
= 2^1/3 x^7/12
4.
72^1/3 x^8/3
(2^3 * 3^2)^1/3 x^8/3
2 (9)^1/3 x^8/3
5.
(3*3*7 *a^5 * b^1 *3^3 * a^6* b^4)^1/2
= (3^5 a^11 b^5)^1/2
= 9 a^5 b^2 (3 a b)^1/2
1 answer