સાદું રૂપ આપો : 

$(i)$ $(3+\sqrt{3})(2+\sqrt{2})$

$(ii)$ $(3+\sqrt{3})(3-\sqrt{3})$

$(iii)$ $(\sqrt{5}+\sqrt{2})^{2}$

$(iv)$ $(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})$

$(i)$ $(3+\sqrt{3})(2+\sqrt{2})=2(3+\sqrt{3})+\sqrt{2}(3+\sqrt{3})$

     $=(2 \times 3+2 \sqrt{3})+(3 \sqrt{2}+\sqrt{2} \times \sqrt{3})$

     $=6+2 \sqrt{3}+3 \sqrt{2}+\sqrt{6}$

$(3+\sqrt{3})(2+\sqrt{2})=6+2 \sqrt{3}+3 \sqrt{2}+\sqrt{6}$

$(ii)$ $(3+\sqrt{3})(3-\sqrt{3})=(3)^{2}-(\sqrt{3})^{2}$

          $=3^{2}-3=9-3=6$ $(3+\sqrt{3})(3-\sqrt{3})=6$

$(iii)$  $(\sqrt{5}+\sqrt{2})^{2}=(\sqrt{5})^{2}+(\sqrt{2})^{2}+2(\sqrt{5})(\sqrt{2})$ 

          $=5+2+2 \sqrt{10}=7+2 \sqrt{10}$ $(\sqrt{5}+\sqrt{2})^{2}=7+2 \sqrt{10}$

$(iv)$  $(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})=(\sqrt{5})^{2}-(\sqrt{2})^{2}$ $=5-2=3$

આમ, $(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})=3$