Избавься от иррациональности в знаменателе дроби: 4-2√х+х/2-√х ​

a) \frac{3}{2 \sqrt{7} } = \frac{3 \times \sqrt{7} }{2 \times \sqrt{7 \times }\sqrt{7} } = \frac{3}{2 \times 7} = \frac{3}{14}a)

2

7

3

=

2×

7×

7

3×

7

=

2×7

3

=

14

3

\begin{gathered}b) \frac{9}{7 + 4 \sqrt{3} } = \frac{9 \times (7 - 4 \sqrt{3} )}{(7 + 4 \sqrt{3} ) \times (7 - 4 \sqrt{3} )} = \frac{63 - 36 \sqrt{3} }{ {7}^{2} -{ (4 \sqrt{3}) }^{2} } = \\ = \frac{63 - 36 \sqrt{3} }{49 - 16 \times 3} = \frac{63 - 36 \sqrt{3} }{49 - 48} = \frac{63 - 36 \sqrt{3} }{1} = 63 - 36 \sqrt{3}\end{gathered}

b)

7+4

3

9

=

(7+4

3

)×(7−4

3

)

9×(7−4

3

)

=

7

2

−(4

3

)

2

63−36

3

=

=

49−16×3

63−36

3

=

49−48

63−36

3

=

1

63−36

3

=63−36

3