Math, asked by dakshasam19, 2 months ago

Please answer this..

Attachments:

Answers

Answered by senboni123456
2

Answer:

Step-by-step explanation:

We have,

\tt{\dfrac{7\sqrt{3} }{\sqrt{10}+\sqrt{3}  }-\dfrac{2\sqrt{5} }{\sqrt{6}+\sqrt{5}  }-\dfrac{3\sqrt{2} }{\sqrt{15}+3\sqrt{2}  }}

\sf{=\dfrac{7\sqrt{3} (\sqrt{10}-\sqrt{3} )}{(\sqrt{10}+\sqrt{3} ) (\sqrt{10}-\sqrt{3} )}-\dfrac{2\sqrt{5}(\sqrt{6}-\sqrt{5} )  }{(\sqrt{6}+\sqrt{5})(\sqrt{6}-\sqrt{5} )  }-\dfrac{3\sqrt{2} (\sqrt{15}-3\sqrt{2} ) }{(\sqrt{15}+3\sqrt{2} )(\sqrt{15}-3\sqrt{2} )  }}

\sf{=\dfrac{7\sqrt{3} (\sqrt{10}-\sqrt{3} )}{10-3 }-\dfrac{2\sqrt{5}(\sqrt{6}-\sqrt{5} )  }{6-5  }-\dfrac{3\sqrt{2} (\sqrt{15}-3\sqrt{2} ) }{15-18  }}

\sf{=\dfrac{7\sqrt{3} (\sqrt{10}-\sqrt{3} )}{7 }-\dfrac{2\sqrt{5}(\sqrt{6}-\sqrt{5} )  }{1 }-\dfrac{3\sqrt{2} (\sqrt{15}-3\sqrt{2} ) }{-3 }}

\sf{=\sqrt{3} (\sqrt{10}-\sqrt{3} )-2\sqrt{5}(\sqrt{6}-\sqrt{5} )  +\sqrt{2} (\sqrt{15}-3\sqrt{2} ) }

\sf{=\sqrt{30}-3-2\sqrt{30}+10  +\sqrt{30}-6 }

\sf{=-3+10  -6 }

\sf{=1}

Similar questions