Question

Rationalize the dinominater:1/{7+3√3}
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Answer 1

Answer:

$\frac{1}{7 + 3 \sqrt{x3} }$ × $\frac{7 - 3\sqrt{3}}{7 - 3\sqrt{3}}$

= $\frac{7 - 3\sqrt{3}}{7^{2} - (3\sqrt{3}) ^{2} }$

= $\frac{7 - 3\sqrt{3}}{49 - (9) (3)}$

= $\frac{7 - 3\sqrt{3}}{49 - 27}$

= $\frac{7 - 3\sqrt{3}}{22}$

HOPE IT HELPS ;)

Answer 2

Answer:

${ \huge{ \sf{ \red{Question:}}}}$

• ${ \sf{Rationalize \: { \frac{1}{7 + 3 \sqrt{3} } }}}$

$\huge{ \sf{ \green{Solution: }}}$

$: { \implies{ \sf{ \frac{1}{7 + 3 \sqrt{3} } }}} \\ \\ : { \implies{ \sf{ \frac{1}{7 + 3 \sqrt{3} } \times \frac{7 - 3 \sqrt{3} }{7 - 3 \sqrt{3} } }}} \\ \\ : { \implies{ \sf{ \frac{7 - 3 \sqrt{3} }{ {(7)}^{2} - (3 { \sqrt{ 3} )}^{2} } }}} \\ \\ : { \implies{ \sf{ \frac{7 - 3 \sqrt{3} }{49 - 9(3)} }}} \\ \\ : { \implies{ \sf{ \frac{7 - 3 \sqrt{3} }{49 - 27} }}} \\ \\ : { \implies{ \sf{ \frac{7 - 3 \sqrt{3} }{22} }}}$