Simplify (the 7th root of x squared)/((1/2)y) times (the cube root of x squared)/(the cube root of y squared) times x/(the cube root of y) times ((9/7) times the square root of y/3)/(5/x).

Step-by-step solution

1. 1

   Convert every radical to a fractional exponent. Using $\sqrt[n]{a^m}=a^{m/n}$: $\sqrt[7]{x^2}=x^{2/7}$, $\sqrt[3]{x^2}=x^{2/3}$, $\sqrt[3]{y^2}=y^{2/3}$, $\sqrt[3]{y}=y^{1/3}$, and $\sqrt{\tfrac y3}=\tfrac{y^{1/2}}{\sqrt3}$. Exponent form is the only practical way to combine roots of different orders.

2. 2

   Clear the fractional coefficients inside the fractions. Dividing by $\tfrac12 y$ means multiplying by $\tfrac{2}{y}$, and dividing by $\tfrac5x$ means multiplying by $\tfrac x5$: $\frac{2x^{2/7}}{y}\times\frac{x^{2/3}}{y^{2/3}}\times\frac{x}{y^{1/3}}\times\frac{9x}{35}\cdot\frac{y^{1/2}}{\sqrt3}.$ Note the fourth factor contributes an extra $x$, which is easy to miss.

3. 3

   Gather the numerical coefficients. $2\times\tfrac{9}{35} = \tfrac{18}{35}$, and the $\tfrac{1}{\sqrt3}$ stays for now: the constant is $\dfrac{18}{35\sqrt3}$.

4. 4

   Add the exponents of x. The powers present are $x^{2/7}$, $x^{2/3}$, $x^1$ (third factor) and $x^1$ (from the fourth): $\frac27+\frac23+1+1 = \frac{6}{21}+\frac{14}{21}+\frac{42}{21} = \frac{62}{21}.$

5. 5

   Add the exponents of y. The numerator supplies $y^{1/2}$; the denominators supply $y^{1}\cdot y^{2/3}\cdot y^{1/3} = y^{2}$. Hence $y^{1/2-2} = y^{-3/2} = \frac{1}{y^{3/2}}.$

6. 6

   Rationalise and reduce the constant. $\dfrac{18}{35\sqrt3} = \dfrac{18\sqrt3}{105} = \dfrac{6\sqrt3}{35}$, so $\text{expression} = \frac{6\sqrt3\,x^{62/21}}{35\,y^{3/2}}.$

7. 7

   Check numerically. At $x=2$, $y=3$ the original product evaluates to $0.44230024$, and $\dfrac{6\sqrt3\cdot 2^{62/21}}{35\cdot 3^{3/2}} = 0.44230024$ - identical to eight decimals.