Question

How do you solve this? Thank you

Answer 1

2)

a)


`\bf a^{\frac{{ n}}{{ m}}} \implies \sqrt[{ m}]{a^( n)} \qquad \qquad \sqrt[{ m}]{a^( n)}\implies a^{\frac{{ n}}{{ m}}}\\\\ -------------------------------\\\\ (4x^5\cdot x^{(1)/(3)})+(2x^4\cdot x^{(1)/(3)})-(7x^3\cdot x^{(1)/(3)})+(3x^2\cdot x^{(1)/(3)})\\\\+(9x^1\cdot x^{(1)/(3)})-(1\cdot x^{(1)/(3)}) \\\\\\ 4x^{5+(1)/(3)}+2x^{4+(1)/(3)}-7x^{3+(1)/(3)}+9x^{1+(1)/(3)}-x^{(1)/(3)}`


`\bf 4x^{(16)/(3)}+2x^{(13)/(3)}-7x^{(10)/(3)}+9x^{(4)/(3)}-x^{(1)/(3)} \\\\\\ 4\sqrt[3]{x^(16)}+2\sqrt[3]{x^(13)}-7\sqrt[3]{x^(10)}+9\sqrt[3]{x^4}-\sqrt[3]{x}`

b)


`\bf \cfrac{4x^5+2x^4-7x^3+3x^2+9x-1}{x^{(1)/(3)}}\impliedby \textit{distributing the denominator} \\\\\\ \cfrac{4x^5}{x^{(1)/(3)}}+\cfrac{2x^4}{x^{(1)/(3)}}-\cfrac{7x^3}{x^{(1)/(3)}}+\cfrac{3x^2}{x^{(1)/(3)}}+\cfrac{9x}{x^{(1)/(3)}}-\cfrac{1}{x^{(1)/(3)}} \\\\\\ (4x^5\cdot x^{-(1)/(3)})+(2x^4\cdot x^{-(1)/(3)})-(7x^3\cdot x^{-(1)/(3)})+(3x^2\cdot x^{-(1)/(3)})\\\\+(9x^1\cdot x^{-(1)/(3)})-(1\cdot x^{-(1)/(3)})`


`\bf 4x^{5-(1)/(3)}+2x^{4-(1)/(3)}-7x^{3-(1)/(3)}+9x^{1-(1)/(3)}-x^{-(1)/(3)} \\\\\\ 4x^{(14)/(3)}+2x^{(11)/(3)}-7x^{(8)/(3)}+9x^{(2)/(3)}-x^{-(1)/(3)} \\\\\\ 4\sqrt[3]{x^(14)}+2\sqrt[3]{x^(11)}-7\sqrt[3]{x^(8)}+9\sqrt[3]{x^(2)}-\frac{1}{\sqrt[3]{x}}`



3)


`\bf \begin{cases} f(x)=√(x)-5x\implies &f(x)x^{(1)/(2)}-5x\\\\ g(x)=5x^2-2x+\sqrt[5]{x}\implies &g(x)=5x^2-2x+x^{(1)/(5)} \end{cases} \\\\\\ \textit{let's multiply the terms from f(x) by each term in g(x)} \\\\\\ x^{(1)/(2)}(5x^2-2x+x^{(1)/(5)})\implies x^{(1)/(2)}5x^2-x^{(1)/(2)}2x+x^{(1)/(2)}x^{(1)/(5)}`


`\bf 5x^{(1)/(2)+2}-2x^{(1)/(2)+1}+x^{(1)/(2)+(1)/(5)}\implies \boxed{5x^{(5)/(2)}-2x^{(3)/(2)}+x^{(7)/(10)}} \\\\\\ -5x(5x^2-2x+x^{(1)/(5)})\implies -5x5x^2-5x2x+5xx^{(1)/(5)} \\\\\\ -25x^3+10x^2-5x^{1+(1)/(5)}\implies \boxed{-25x^3+10x^2-5x^{(6)/(5)}}`


`\bf 5√(x^5)-2√(x^3)+\sqrt[10]{x^7}-25x^3+10x^2-5\sqrt[5]{x^6}`