Tentukan hasil perkalian dan pembagian bentuk akar berikut A.√5×√2 B.2√13×√9 C.√6(5+√12) D.(√2+√3)(√2-√3) E.√18/√2 F.√20/2√5 G.√24/√45×√9/√2 H.7√32/6√28×7√27

[tex]a.\ \sqrt{5} \times \sqrt{2} = \sqrt{5 \times 2} = \sqrt{10} \\ \\ b.\ 2 \sqrt{13} \times \sqrt{9}= 2 \times \sqrt{13} \times 3=6\sqrt{13} \\ \\ c.\ \sqrt{6}.(5+ \sqrt{12})=\sqrt{6}.(5+ \sqrt{6 \times 2}) \\ \\ .\ =5 \sqrt{6}+( \sqrt{6} \times \sqrt{6} \times \sqrt{2} ) =5 \sqrt{6}+6\sqrt{2} \\ \\ d.\ ( \sqrt{2}+ \sqrt{3})( \sqrt{2}- \sqrt{3}).........Ingat\ bahwa\ (a + b)(a - b)=a^2-b^2 \\ \\ .\ = ( \sqrt{2})^2-( \sqrt{3})^2=2-3=-1 [/tex]

[tex]e.\ \frac{ \sqrt{18} }{ \sqrt{2} }= \frac{ \sqrt{9 \times 2} }{ \sqrt{2} }=\frac{ \sqrt{9} \times \sqrt{2} }{ \sqrt{2} }=\frac{ 3 \times \sqrt{2} }{ \sqrt{2} }=3 \\ \\ f.\ \frac{ \sqrt{20} }{ 2\sqrt{5} }= \frac{ \sqrt{4 \times 5} }{ 2\sqrt{5} }=\frac{ \sqrt{4} \times \sqrt{5} }{ 2\sqrt{5} }=\frac{ 2 \sqrt{5} }{ 2\sqrt{5} }=1 [/tex]

[tex]g.\ \frac{\sqrt{24}}{ \sqrt{45}} \times \frac{\sqrt{9} }{\sqrt{2}}=\frac{\sqrt{4 \times 3\times 2}}{\sqrt{9 \times 5}} \times \frac{3}{\sqrt{2}}=\frac{\sqrt{4} \times \sqrt{3} \times \sqrt{2}}{\sqrt{9} \times \sqrt{5}} \times \frac{3}{\sqrt{2}}=\frac{2 \times \sqrt{3} \times \sqrt{2}}{3\times \sqrt{5}} \times \frac{3}{\sqrt{2}} \\ \\ .\ =\frac{2\sqrt{3}}{\sqrt{5}} [/tex]

[tex].\ =\frac{ 2\sqrt{3}}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}}=\frac{ 2\sqrt{3} \times \sqrt{5}}{\sqrt{5} \times \sqrt{5}} = \frac{2 \sqrt{15}}{5} = \frac{2}{5} \sqrt{15} \\ \\ h.\ \frac{7 \sqrt{32}}{6 \sqrt{28} \times 7 \sqrt{27}} =\frac{7 \times \sqrt{16} \times \sqrt{2}}{6 \times \sqrt{4} \times \sqrt{7} \times 7 \times \sqrt{9} \times \sqrt{3}}=\frac{7 \times 4 \times \sqrt{2}}{6 \times 2 \times \sqrt{7} \times 7 \times 3 \times \sqrt{3}}= \frac{\sqrt{2}}{9 \sqrt{21}} [/tex]

[tex]= \frac{\sqrt{2}}{9 \sqrt{21}} \times \frac{ \sqrt{21}}{ \sqrt{21}}= \frac{\sqrt{2} \sqrt{21} }{9 \sqrt{21} \sqrt{21} }= \frac{\sqrt{42} }{9 \times 21}= \frac{1}{189} \sqrt{42} [/tex]