Solution:

\(\frac{2}{3√5}\) × \(\frac{√5}{√5}\)

= \(\frac{2√5}{3×5}\)

= \(\frac{2√5}{15}\)

Solution:

\(\frac{3}{√2}\)×\(\frac{√2}{√2}\)

=\(\frac{3√2}{√2^2}\)

=\(\frac{3√2}{2}\)

Solution:

\(\frac{5+√3}{√5}\)×\(\frac{√5}{√5}\)

=\(\frac{5√5+√15}{5}\)

Solution:

3√5+6√5

=(3+6)√5

=9√5

Solution:

3√10 - 3√10

= (3-3)√10

= 0×√10

= 0

Solution:

3√20+2√45

= 3\(\sqrt{2×2×5}\)+2\(\sqrt{3×3×5}\)

= 3\(\sqrt{2^2×5}\)+2\(\sqrt{3^2×5}\)

= 3×2√5+2×3√5

= 6√5+6√5

= (6+6)√5

= 12√5

Solution:

(5√7 × 3√5) × 4√3

= 15\(\sqrt{7×5}\) × 4√3

= 15√35 × 4√3

= 60\(\sqrt{35×3}\)

= 60√105

Solution:

(2√3 × 3√5) + 5√15

= (6\(\sqrt{3×5}\) + 5√15

= (6\(\sqrt{15}\) + 5√15

= (6+5)√15

=11√15

Solution:

√125-√45

= \(\sqrt{25×5}\) - \(\sqrt{9×5}\)

= 5√5 - 3√5

= 2√5

Solution:

3√2 - 4√2 + 5√2

= 3\(\sqrt{2}\) + 5\(\sqrt{2}\) - 4√2

= 8√2 - 4√2

= 4√2

Solution:

\(\sqrt{128}\) - \(\sqrt{50}\)

= \(\sqrt{2\times2\times2\times2\times2\times2\times2}\) - \(\sqrt{2\times5\times5}\)

= \(\sqrt{2^2\times2^2\times2^2\times2}\) - \(\sqrt{2\times5^2}\)

= 2\(\times\)2\(\times\)2\(\sqrt{2}\) - 5\(\sqrt{2}\)

= 8\(\sqrt{2}\) - 5\(\sqrt{2}\)

= (8 - 5) \(\sqrt{2}\)

= 3\(\sqrt{2}\)

Solution:

\(\frac{√6+√10}{√2}\)

=\(\frac{√6+√10}{√2}\)×\(\frac{√2}{√2}\)

= \(\frac{√2(√6+√10)}{2}\)

= \(\frac{(√2+√20)}{√2}\)

= \(\frac{√4×3+√4×5}{2}\)

= \(\frac{2√3+2√5}{2}\)

= \(\frac{2(√3+√5)}{2}\)

= √3+√5

Soutioln:

√288 - √72 + √8

= \(\sqrt{2×2×2×2×2×3×3}\) - \(\sqrt{2×2×2×3×3}\) + \(\sqrt{2×2×2}\)

= \(\sqrt{2^2 × 2^2 × 2 × 3^2}\) - \(\sqrt{2^2× 2 × 3^2}\) + \(\sqrt{2^2 ×2}\)

= 2 × 2 × 3√2 - 2 × 3√2 + 2√2

= (12-6+2)√2

= (14-6)√2

= 8√2

Solution:

√128-√50

= \(\sqrt{2×2×2×2×2×2×2}\)-\(\sqrt{2×5^2}\)

=\(\sqrt{2^2×2^2×2^2×2}\)-\(\sqrt{2×5^2}\)

= 2 × 2 × 2√2 - 5√2

= 8√2 - 5√2

= (8-5)√2

= 3√2

Solution:

\(\frac{3}{√2}\)+ 5

= \(\frac{3×√2}{√2×√2}\)+ 5

= \(\frac{3√2}{(√2)^2}\)+ 5

= \(\frac{3√2}{2}\)+ 5

= \(\frac{3×√2+10}{2}\)

Solution:

\(\frac{2}{√5}\)+\(\frac{3}{√2}\)

=\(\frac{2}{√5}\)×\(\frac{√5}{√5}\)+\(\frac{3}{√2}\)+\(\frac{√2}{√2}\) (Rationalise the denominator of each term)

=\(\frac{2√5}{5}\)+\(\frac{3√2}{2}\)

=\(\frac{4√5+15√2}{10}\) (LCM of 5 and 2 = 10)

Solution:

\(\sqrt{63}\) - 2\(\sqrt{28}\) + 5 \(\sqrt{7}\)

= \(\sqrt{3\times3\times7}\) - 2\(\sqrt{2\times2\times7}\) + 5\(\sqrt{7}\)

=\(\sqrt{3^2\times7}\) - 2\(\sqrt{2^2\times7}\) + 5\(\sqrt{7}\)

= 3\(\sqrt{7}\) - 2\(\times\)2\(\sqrt{7}\) + 5\(\sqrt{7}\)

= 3 \(\sqrt{7}\) - 4\(\sqrt{7}\) + 5\(\sqrt{7}\)

= (3-4+5)\(\sqrt{7}\)

= (8-4)\(\sqrt{7}\)

= 4\(\sqrt{7}\)

Solution:

(3 \(\sqrt{7}\) + 2\(\sqrt{28}\) \(\times\) 4\(\sqrt{7}\)

=( 3 \(\sqrt{7}\) + 2 \(\sqrt{2\times2\times7}\)) \(\times\) 4\(\sqrt{7}\)

= (3 \(\sqrt{7}\) + 2\(\times\)2\(\sqrt{7}\) \(\times\) 4\(\sqrt{7}\)

= (3\(\sqrt{7}\) + 4\(\sqrt{7}\) ) \(\times\) 4\(\sqrt{7}\)

= (3+4) \(\sqrt{7}\) \(\times\) 4\(\sqrt{7}\)

= 7 \(\sqrt{7}\) \(\times\) 4\(\sqrt{7}\)

= 28\(\sqrt{7\times7}\)

= 28\(\sqrt{7^2}\)

= 28\(\times\)7

= 196

Solution:

21\(\sqrt{7}\) - 3\(\sqrt{28}\) + \(\sqrt{63}\)

= 21\(\sqrt{7}\) - 3\(\sqrt{2\times2\times7}\) + \(\sqrt{3\times3\times7}\)

= 21\(\sqrt{7}\) - 3\(\sqrt{2^2\times7}\) + \(\sqrt{3^2\times7}\)

= 21\(\sqrt{7}\) - 3\(\times\)2\(\sqrt{7}\) + 3\(\sqrt{7}\)

= 21\(\sqrt{7}\) - 6\(\sqrt{7}\) + 3\(\sqrt{7}\)

= (21 - 6 + 3)\(\sqrt{7}\)

= 18\(\sqrt{7}\)

Solution:

3\(\sqrt{20}\) + 2 \(\sqrt{45}\)

= 3\(\sqrt{2\times2\times5}\) + 2\(\sqrt{3\times3\times5}\)

= 3\(\times\)2 \(\sqrt{5}\) + 2\(\times\)3\(\sqrt{5}\)

= 3\(\sqrt{2^2\times5}\) + 2\(\sqrt{3^2\times5}\)

= 3\(\times\)2\(\sqrt{5}\) + 2\(\times\)3\(\sqrt{5}\)

= 6\(\sqrt{5}\) + 6\(\sqrt{5}\)

= (6+6)\(\sqrt{5}\)

= 12\(\sqrt{5}\)