Let's simplify each of the given expressions step-by-step:
### i) [tex]\(\frac{7^6}{7^4}\)[/tex]
We can use the property of exponents that states [tex]\(\frac{a^m}{a^n} = a^{m-n}\)[/tex].
[tex]\[
\frac{7^6}{7^4} = 7^{6-4} = 7^2 = 49
\][/tex]
### ii) [tex]\(\frac{2^4 \cdot 5^3}{10^2}\)[/tex]
First, express [tex]\(10^2\)[/tex] in terms of its prime factors:
[tex]\[
10^2 = (2 \cdot 5)^2 = 2^2 \cdot 5^2
\][/tex]
Now substitute this back into the expression:
[tex]\[
\frac{2^4 \cdot 5^3}{2^2 \cdot 5^2}
\][/tex]
Apply the property [tex]\(\frac{a^m}{a^n} = a^{m-n}\)[/tex]:
[tex]\[
= \frac{2^4}{2^2} \cdot \frac{5^3}{5^2} = 2^{4-2} \cdot 5^{3-2} = 2^2 \cdot 5 = 4 \cdot 5 = 20
\][/tex]
### iii) [tex]\(\left\{\frac{(a+b)^2 \cdot (c+d)^3}{(a+b) \cdot (c+d)^2}\right\}^3\)[/tex]
First, simplify the fraction inside the braces:
[tex]\[
\frac{(a+b)^2 \cdot (c+d)^3}{(a+b) \cdot (c+d)^2} = (a+b)^{2-1} \cdot (c+d)^{3-2} = (a+b) \cdot (c+d)
\][/tex]
Now raise this result to the power of 3:
[tex]\[
\left\{(a+b) \cdot (c+d)\right\}^3 = (a+b)^3 \cdot (c+d)^3
\][/tex]
### iv) [tex]\((\sqrt[3]{a})^{\frac{1}{2}}\)[/tex]
Rewrite the cube root and then apply the power rule:
[tex]\[
(\sqrt[3]{a})^{\frac{1}{2}} = (a^{\frac{1}{3}})^{\frac{1}{2}} = a^{\frac{1}{3} \cdot \frac{1}{2}} = a^{\frac{1}{6}}
\][/tex]
### v) [tex]\(\sqrt[5]{x^5} \cdot \sqrt[4]{x^4}\)[/tex]
Rewrite the roots and then simplify:
[tex]\[
\sqrt[5]{x^5} = (x^5)^{\frac{1}{5}} = x^{5 \cdot \frac{1}{5}} = x^1 = x
\][/tex]
[tex]\[
\sqrt[4]{x^4} = (x^4)^{\frac{1}{4}} = x^{4 \cdot \frac{1}{4}} = x^1 = x
\][/tex]
Now multiply these results:
[tex]\[
\sqrt[5]{x^5} \cdot \sqrt[4]{x^4} = x \cdot x = x^2
\][/tex]
### Summary:
i) [tex]\(\frac{7^6}{7^4} = 49\)[/tex]
ii) [tex]\(\frac{2^4 \cdot 5^3}{10^2} = 20\)[/tex]
iii) [tex]\(\left\{\frac{(a+b)^2 \cdot (c+d)^3}{(a+b) \cdot (c+d)^2}\right\}^3 = (a+b)^3 \cdot (c+d)^3\)[/tex]
iv) [tex]\((\sqrt[3]{a})^{\frac{1}{2}} = a^{\frac{1}{6}}\)[/tex]
v) [tex]\(\sqrt[5]{x^5} \cdot \sqrt[4]{x^4} = x^2\)[/tex]
