Answer
$$\frac{4}{7+3\sqrt{2}}=\frac{28-12\sqrt{2}}{31}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{4}{7+3\sqrt{2}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{4}{7+3\sqrt{2}}\frac{7-3\sqrt{2}}{7-3\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{28-12\sqrt{2}}{49-21\sqrt{2}+21\sqrt{2}-18} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{28-12\sqrt{2}}{31}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 7- 3 \sqrt{2}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 4 } \cdot \left( 7- 3 \sqrt{2}\right) = \color{blue}{4} \cdot7+\color{blue}{4} \cdot- 3 \sqrt{2} = \\ = 28- 12 \sqrt{2} $$ Simplify denominator. $$ \color{blue}{ \left( 7 + 3 \sqrt{2}\right) } \cdot \left( 7- 3 \sqrt{2}\right) = \color{blue}{7} \cdot7+\color{blue}{7} \cdot- 3 \sqrt{2}+\color{blue}{ 3 \sqrt{2}} \cdot7+\color{blue}{ 3 \sqrt{2}} \cdot- 3 \sqrt{2} = \\ = 49- 21 \sqrt{2} + 21 \sqrt{2}-18 $$ |
| ③ | Simplify numerator and denominator |
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