Respuesta :

[tex] \tt \huge \leadsto \frac{ \sqrt{7} - \sqrt{3} }{ \sqrt{7} + \sqrt{3} } [/tex]

[tex] \tt \huge \leadsto \frac{ \sqrt{7} - \sqrt{3} }{ \sqrt{7} + \sqrt{3}} \times \frac{ \sqrt{7} - \sqrt{3} }{ \sqrt{7} - \sqrt{3} } [/tex]

[tex] \tt \huge \leadsto \frac{7 - 3}{ (\sqrt{7 })^{2} - ( \sqrt{3})^{2} } [/tex]

[tex] \tt \huge \leadsto \frac{4}{7 - 3} [/tex]

[tex] \tt \huge \leadsto\frac{4}{4} [/tex]

[tex]\tt\huge\leadsto{1}[/tex]

devishri1977

Answer:

Step-by-step explanation:

To rationalize the denominator multiply the numerator and denominator by the conjugate of √7 + √3 = √7- √3

[tex]\frac{\sqrt{7}-\sqrt{3}}{\sqrt{7}+\sqrt{3}}=\frac{(\sqrt{7}-\sqrt{3})(\sqrt{7}-\sqrt{3})}{(\sqrt{7}+\sqrt{3})(\sqrt{7}-\sqrt{3})}\\\\\\= \frac{(\sqrt{7}-\sqrt{3})^{2}}{(\sqrt{7})^{2}-(\sqrt{3})^{2}}\\\\\\= \frac{(\sqrt{7})^{2}-2*(\sqrt{7})*(\sqrt{3})+(\sqrt{3})^{2})}{7-3}\\\\=\frac{7-2\sqrt{21}+3}{4}\\\\=\frac{10-2\sqrt{21}}{4}\\\\=\frac{2(5-\sqrt{21})}{4}\\\\=\frac{5-\sqrt{21}}{2}[/tex]

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