Question

15 for this question, note this example: \\(\frac{2a^3}{4a} = \frac{2_1 \times a \times a \times a_1}{2_4 \times a_1} = \frac{a^2}{2}\\)

simplify these expressions with higher powers.

a \\(\frac{a^4}{a}\\)

b \\(\frac{3b^3}{9b}\\)

c \\(\frac{4ab^3}{12ab}\\)

e \\(-\frac{2a^5}{3a^2}\\)

f \\(-\frac{8a^5}{20a^7}\\)

g \\(\frac{4a^3}{10a^8}\\)

i \\(\frac{15a^4b^2}{5ab}\\)

j \\(\frac{28a^3b^5}{7a^4b^2}\\)

k \\(\frac{2a^5b^2}{6a^2b^3}\\)

Explanation:

Simplify parts a, b, and c

Using the Cancelling Common Factors and Algebraic Simplification knowledge points

$$
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Simplify parts e, f, and g

Using the Cancelling Common Factors and Algebraic Simplification knowledge points

$$
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Simplify parts i, j, and k

Using the Cancelling Common Factors and Algebraic Simplification knowledge points

$$
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$$

Answer:

No.	Answer
b	\(\frac{b^2}{3}\)
c	\(\frac{b^2}{3}\)
e	\(-\frac{2a^3}{3}\)
f	\(-\frac{2}{5a^2}\)
g	\(\frac{2}{5a^5}\)
i	\(3a^3b\)
j	\(\frac{4b^3}{a}\)
k	\(\frac{a^3}{3b}\)