Question

Question 9
Simplify the following
$\left(i\right)\sqrt{45}-3\sqrt{20}+4\sqrt{5}\left(ii\right)\frac{\sqrt{24}}{8}+\frac{\sqrt{54}}{9}\left(iii\right)4\sqrt{12}\times 7\sqrt{6}\left(iv\right)4\sqrt{28}+3\sqrt{7}+\sqrt[3]{37}\left(v\right)3\sqrt{3}+2\sqrt{27}+\frac{7}{\sqrt{3}}\left(vi\right)(\sqrt{3}-\sqrt{2}{)}^2(vii)\sqrt[4]{481}-8\sqrt[3]{3216}+15\sqrt[5]{532}+\sqrt{225}(viii)\frac{3}{\sqrt{8}}+\frac{1}{\sqrt{2}}(ix)\frac{2\sqrt{3}}{3}-\frac{\sqrt{3}}{6}$

Answer 1

$\left(i\right)\sqrt{45}-3\sqrt{20}+4\sqrt{5}=\sqrt{3\times 3\times 5}-3\sqrt{2\times 2\times 5}+4\sqrt{5}=3\sqrt{5}-3\times 2\sqrt{5}+4\sqrt{5}=3\sqrt{5}-6\sqrt{5}+4\sqrt{5}=\sqrt{5}$

$\left(ii\right)\frac{\sqrt{24}}{8}+\frac{\sqrt{54}}{9}=\frac{\sqrt{2\times 2\times 2\times 3}}{8}+\frac{\sqrt{3\times 3\times 3\times 2}}{9}=\frac{2\sqrt{6}}{8}+\frac{3\sqrt{6}}{9}=\frac{\sqrt{6}}{4}+\frac{\sqrt{6}}{3}=\frac{3\sqrt{6}+4\sqrt{6}}{12}=\frac{7\sqrt{6}}{12}$

$\left(iii\right)4\sqrt{12}\times 7\sqrt{6}=4\sqrt{2\times 2\times 3}\times 7\sqrt{2\times 6}$
$=8\sqrt{3}\times 7\sqrt{2\times 3}$
$=168\sqrt{2}$


$\left(iv\right)4\sqrt{28}÷3\sqrt{7}÷\sqrt[3]{37}=\frac{8}{3\times 7^{\frac{1}{3}}}$

$\left(v\right)3\sqrt{3}+2\sqrt{27}+\frac{7}{\sqrt{3}}=3\sqrt{3}+2\sqrt{3\times 3\times 3}+\frac{7}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=3\sqrt{3}+6\sqrt{3}+\frac{7\sqrt{3}}{3}=9\sqrt{3}+\frac{7\sqrt{3}}{3}$
$\frac{27\sqrt{3}+7\sqrt{3}}{3}=\frac{34\sqrt{3}}{3}$
$\left(vi\right)(\sqrt{3}-\sqrt{2}{)}^2=(\sqrt{3}{)}^2+(\sqrt{2}{)}^2-2\sqrt{3}\times \sqrt{2}[Usingidentity(a-b{)}^2=a^2+b^2-2ab]=3+2-2\sqrt{3\times 2}=5-2\sqrt{6}$

$\left(vii\right)\sqrt[4]{481}-8\sqrt[3]{3216}+15\sqrt[5]{532}+\sqrt{225}=(81{)}^{\frac{1}{4}}-8\times (216{)}^{\frac{1}{3}}+15\times (32{)}^{\frac{1}{5}}+\sqrt{(15{)}^2}[\therefore \sqrt[m]{ma}=a^{1/m}]$
$=(3^4{)}^{\frac{1}{4}}-8\times 6^{3.\frac{1}{3}}+15\times 2^{5.\frac{1}{5}}+15$
$=3^{4.\frac{1}{4}}-8\times 6^{3.\frac{1}{3}}+15\times 2^{5.\frac{1}{5}}+15[\therefore (a^m{)}^n=a^{mn}]$
$=3^1-8\times 6^1+15\times 2^1+15$
$=3-48+30+15$
$=48-48=0$

$\left(viii\right)\frac{3}{\sqrt{8}}+\frac{1}{\sqrt{2}}=\frac{3}{\sqrt{2\times 2\times 2}}+\frac{1}{\sqrt{2}}=\frac{3}{2\sqrt{2}}+\frac{1}{\sqrt{2}}$
$=\frac{3+2}{2\sqrt{2}}=\frac{5}{2\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}$
[Multiplying numerator and denominator by $\sqrt{2}$ ]
$=\frac{5\sqrt{2}}{2\times 2}=\frac{5\sqrt{2}}{4}$
$\left(ix\right)\frac{2\sqrt{3}}{3}-\frac{\sqrt{3}}{6}=\frac{4\sqrt{3}-\sqrt{3}}{6}=\frac{3\sqrt{3}}{6}=\frac{\sqrt{3}}{2}$