Question

Simplify by rationalizing the denominator.

(i) $\frac{4}{7}\sqrt{147}+\frac{3}{8}\sqrt{192}-\frac{1}{5}\sqrt{75}$

(ii) $5\sqrt{3}+2\sqrt{27}+\frac{1}{\sqrt{3}}$

(iii) $\sqrt{216}-5\sqrt{6}+\sqrt{294}-\frac{3}{\sqrt{6}}$

(iv) $\sqrt{{\left(x+y\right)}^3}-\sqrt{x^3+x^{2}y}-\sqrt{x{y}^2+y^3}$

(v) $4\sqrt{12}-\sqrt{75}-7\sqrt{48}$

(vi) $2\sqrt{48}-\sqrt{75}-\frac{1}{\sqrt{3}}$

(vii) $\sqrt[3]{2}+\sqrt[3]{16}-\sqrt[3]{54}$

(viii) $7\sqrt[3]{40}+3\sqrt[3]{625}-4\sqrt[3]{320}$

(ix) $\sqrt[4]{162}+7\sqrt[4]{32}-\sqrt[4]{1250}$

(x) $\sqrt[4]{81}-8\sqrt[3]{216}+15\sqrt[5]{32}+\sqrt{225}$

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\frac{4}{7}\sqrt{147}+\frac{3}{8}\sqrt{192}-\frac{1}{5}\sqrt{75} \\ =\frac{4}{7}\sqrt{49\times 3}+\frac{3}{8}\sqrt{64\times 3}-\frac{1}{5}\sqrt{25\times 3} \\ =\frac{4}{7}\times 7\sqrt{3}+\frac{3}{8}\times 8\sqrt{3}-\frac{1}{5}\times 5\sqrt{3} \\ =4\sqrt{3}+3\sqrt{3}-\sqrt{3} \\ =6\sqrt{3} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)5\sqrt{3}+2\sqrt{27}+\frac{1}{\sqrt{3}} \\ =5\sqrt{3}+2\sqrt{9\times 3}+\frac{1}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}} \\ =5\sqrt{3}+6\sqrt{3}+\frac{\sqrt{3}}{3} \\ =\left(5+6+\frac{1}{3}\right)\sqrt{3} \\ =\frac{34}{3}\sqrt{3} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right)\sqrt{216}-5\sqrt{6}+\sqrt{294}-\frac{3}{\sqrt{6}} \\ =\sqrt{36\times 6}-5\sqrt{6}+\sqrt{49\times 6}-\frac{3}{\sqrt{6}}\times \frac{\sqrt{6}}{\sqrt{6}} \\ =6\sqrt{6}-5\sqrt{5}+7\sqrt{6}-\frac{1}{2}\sqrt{6} \\ =\left(6-5+7-\frac{1}{2}\right)\sqrt{6} \\ =\left(8-\frac{1}{2}\right)\sqrt{6} \\ =\frac{15\sqrt{6}}{2} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iv}\right)\sqrt{{\left(x+y\right)}^3}-\sqrt{x^3+x^{2}y}-\sqrt{x{y}^2+y^3} \\ =\sqrt{{\left(x+y\right)}^2\left(x+y\right)}-\sqrt{x^2\left(x+y\right)}-\sqrt{y^2\left(x+y\right)} \\ =\left(x+y\right)\sqrt{x+y}-x\sqrt{x+y}-y\sqrt{x+y} \\ =\left(x+y-x-y\right)\sqrt{x+y} \\ =0 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{v}\right)4\sqrt{12}-\sqrt{75}-7\sqrt{48} \\ =4\sqrt{4\times 3}-\sqrt{25\times 3}-7\sqrt{16\times 3} \\ =8\sqrt{3}-5\sqrt{3}-28\sqrt{3} \\ =\left(8-5-28\right)\sqrt{3} \\ =-25\sqrt{3} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{vi}\right)2\sqrt{48}-\sqrt{75}-\frac{1}{\sqrt{3}} \\ =2\sqrt{16\times 3}-5\sqrt{25\times 3}-\frac{1}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}} \\ =8\sqrt{3}-5\sqrt{3}-\frac{\sqrt{3}}{3} \\ =\left(8-5-\frac{1}{3}\right)\sqrt{3} \\ =\frac{8}{3}\sqrt{3} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{vii}\right)\sqrt[3]{2}+\sqrt[3]{16}-\sqrt[3]{54} \\ =\sqrt[3]{2}+\sqrt[3]{2^3\times 2}-\sqrt[3]{3^3\times 2} \\ =\sqrt[3]{2}+2\sqrt[3]{2}-3\sqrt[3]{2} \\ =\left(1+2-3\right)\sqrt[3]{2} \\ =0 \end{gathered}$$


$$\begin{gathered} \left(\mathrm{viii}\right)7\sqrt[3]{40}+3\sqrt[3]{625}-4\sqrt[3]{320} \\ =7\sqrt[3]{2^3\times 5}+3\sqrt[3]{5^3\times 5}-4\sqrt[3]{4^3\times 5} \\ =14\sqrt[3]{5}+15\sqrt[3]{5}-16\sqrt[3]{5} \\ =\left(14+15-16\right)\sqrt[3]{5} \\ =13\sqrt[3]{5} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ix}\right)\sqrt[4]{162}+7\sqrt[4]{32}-\sqrt[4]{1250} \\ =\sqrt[4]{3^4\times 2}+7\sqrt[4]{2^4\times 2}-\sqrt[4]{5^4\times 2} \\ =3\sqrt[4]{2}+14\sqrt[4]{2}-5\sqrt[4]{2} \\ =\left(3+14-5\right)\sqrt[4]{2} \\ =12\sqrt[4]{2} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{x}\right)\sqrt[4]{81}-8\sqrt[3]{216}+15\sqrt[5]{32}+\sqrt{225} \\ =\sqrt[4]{3^4}-8\sqrt[3]{6^3}+15\sqrt[5]{2^5}+\sqrt{{15}^2} \\ =3-48+30+15 \\ =0 \end{gathered}$$