Math, asked by sarahprakash34, 1 month ago

Simplify by rationalizing denominators.\frac{2\sqrt{3}-\sqrt2 }{2\sqrt{3}+\sqrt{2} }

Answers

Answered by xSoyaibImtiazAhmedx
2

 \tt \bold{\frac{2\sqrt{3}-\sqrt2 }{2\sqrt{3}+\sqrt{2}}}

 =  \frac{(2\sqrt{3}-\sqrt2 )(2 \sqrt{3}  -  \sqrt{2}) }{(2\sqrt{3}+\sqrt{2})(2\sqrt{3} - \sqrt{2})}

=  \frac{(2\sqrt{3}-\sqrt2 ) ^{2}  }{(2\sqrt{3} )^{2}  - (\sqrt{2} )^{2}}

=  \frac{(2 \sqrt{3})^{2}  -2 \times  2 \sqrt{3}  + (  \sqrt{2})^{2}    }{4 \times 3  - 2}

=  \frac{4 \times 3  -4 \sqrt{3}  +   {2}    }{12  - 2}

=  \frac{12 -4 \sqrt{3}  + 2  }{10}

=  \frac{14 -4 \sqrt{3}   }{10}

=  \frac{2(7-2 \sqrt{3}  ) }{2×5}

 =   \boxed{\frac{7 - 2 \sqrt{3} }{5} }

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