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Simplest rationalizing factor of 2√5 - √3

= 1/(2√5 - √3)

= 2√5 + √3

Simplest rationalizing factor of √3 +√5

1/(√3 +√5) = √3 - √5

(√8 – √2 ) (√8 + √2) 

= (√8)² – (√2)²  [(a+b)(a-b) = a² - b²] 

= 8 - 2

= 6

(i)  \(\sqrt[3]{4}\) x \(\sqrt[3]{16}\)

= \(\sqrt[3]{4\times 16}\) = \(\sqrt[3]{64}\)

= \(\sqrt[3]{4^3}\) = \((4^3)^\frac{1}{3}\) = 4

(ii) \(\frac{\sqrt[4]{1250}}{\sqrt[4]{2}}\) = \(\sqrt[4]\frac{1250}{2}\)

= \(\sqrt[4]\frac{625\times 2}{2}\) = \(\sqrt[4]{625}\)

= \(\sqrt[4]{5^4}\) = (5⁴)^{\(\frac{1}{4}\)} = 5

(3 + √3)(5- √2 ) 

= 15 – 3√2 + 5√3 – √6

(3 + √3) (3 – √3) 

= (3)² – (√3)²   [(a+b)(a-b) = a² - b²] 

= 9 – 3 

= 6

= (4 + √7) (3 + √2)

= 12 + 4√2 + 3√7 + √14

(√5 – √2) (√5 + √2) 

= (√5)² – (√2)²   [(a+b)(a-b) = a² - b²] 

= 5 – 2 

= 3

(√3 + √7)² 

= (√3)² + (√7)² + 2(√3)( √7)  [(a + b)2 = a2 + b2 + 2ab]

= 3 + 7 + 2√21 

= 10 + 2√21

(√5 – √3)² 

= (√5)² + (√3)² – 2(√5)( √3)  [(a – b)2 = a2 + b2 – 2ab]

= 5 + 3 – 2√15 

= 8 – 2√15

(2√5 + 3√2)² 

= (2√5)² + (3√2)² + 2(2√5)(3√2) [(a + b)2 = a2 + b2 + 2ab]

= 20 + 18 + 12√10 

= 38 + 12√10

Given x +√15 = 4

x = 4 - √15

\(\frac{1}{x}\) = 1/(4 - √15) = (4 + √15) / 16 -15 = 4 + √15

So, x + \(\frac{1}{x}\) = 4 - √15 + 4+√15 = 8

Given, x = 7 + 4√3 , xy = 1

Y = \(\frac{1}{x}\) = \(\frac{1}{7}\)+ 4√3 = 7 - 4√3

Y² = \(\frac{1}{x^2}\) = 49 + 48 - 56√3 = 97 - 56√3

Similarly, x = \(\frac{1}{y}\)

= x² = \(\frac{1}{y^2}\) = ( 7 + 4√3)² = 49 + 48 + 56√3 = 97+ 56√3

So, \(\frac{1}{x^2}\) + \(\frac{1}{y^2}\) = 97 + 56√3 + 97 – 56√3 = 194

Given x = \(\sqrt[3]{2+\sqrt3}\)

= x³ = 2 + √3

Similarly, \(\frac{1}{x^3}\) = 2 - √3

X³ + \(\frac{1}{x^3}\) = 2 +√3 + 2 - √3 = 4.

\(x = \sqrt2-1\)

\(\frac{1}{x} = \frac{1}{(\sqrt2-1)}\)

Rationalising denominator, we have

\(= \frac{1}{(\sqrt2-1)} \) x \(\frac {(\sqrt2+1)}{(\sqrt2+1)}\)

\(= \frac{(\sqrt2+1)}{(2-1)}\)

\(= \sqrt2+1\)

(i) Multiply by √3 - √2 both numerator and denominator.

\(\frac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}\)

\(= {(\sqrt3-\sqrt2)(\sqrt3-\sqrt2)\over (\sqrt3+\sqrt2)(\sqrt3-\sqrt2)}\)

\(= {(\sqrt3-\sqrt2)^2\over3-2}\)

\(=\frac {3-2\sqrt3\sqrt2+2}{1}\)

\(= 5-2\sqrt6\)

(ii)  Multiply by 7 - 4√3 both numerator and denominator.

 \(\frac{5+2\sqrt3}{7+4\sqrt3}\)

\(= \frac{(5+2\sqrt3)(7-4\sqrt3)} {(7+4\sqrt3)(7-4\sqrt3)}\)

\(= \frac{(5+2\sqrt3)(7-4\sqrt3)} {49-48}\)

\(= 36-20\sqrt3 + 14\sqrt3 - 24\)

\(= 11-6\sqrt3\)

(iii) Multiply by 3+2√2 both numerator and denominator.

 \(\frac{1+\sqrt2}{3-2\sqrt2}\)

\(= \frac{(1+\sqrt2)(3+2\sqrt2)}{(3-2\sqrt2)(3+2\sqrt2)}\)

 \(= \frac{(1+\sqrt2)(3+2\sqrt2)}{9-8}\)

\(= 3 +2\sqrt2 + 3\sqrt2 + 4\)

\(= 7 + 5\sqrt2\)

(iv) Multiply by 3√5 + 2√6 both numerator and denominator.

 \(\frac{2\sqrt6-\sqrt5}{3\sqrt5-2\sqrt6}\)

\(= \frac{(2\sqrt6 - \sqrt5)(3\sqrt5+2\sqrt6)}{(3\sqrt5-2\sqrt6)(3\sqrt5+2\sqrt6)}\)

\(= \frac{(2\sqrt6 - \sqrt5)(3\sqrt5+2\sqrt6)}{45-24}\)

\(= \frac{(2\sqrt6 - \sqrt5)(3\sqrt5+2\sqrt6)}{21}\)

\(= \frac{6\sqrt{30}+24-15-2\sqrt{30}}{21}\)

\(= \frac{4\sqrt{30}+9}{21}\)

(v) Multiply by √48 - √18 both numerator and denominator.

 \(\frac{4\sqrt3+5\sqrt2}{\sqrt{48}+\sqrt{18}}\)

\(= \frac{(4\sqrt3+5\sqrt2)(\sqrt{48}-\sqrt{18})}{(\sqrt{48}+\sqrt{18})(\sqrt{48}-\sqrt{18})}\)

\(= \frac{(4\sqrt3+5\sqrt2)(\sqrt{48}-\sqrt{18})}{48-18}\)

\(= \frac{48-12\sqrt6+20\sqrt6-30}{30}\)

\(=\frac{18+8\sqrt6}{30}\)

\(=\frac{9+4\sqrt6}{15}\)

(vi) Multiply by 2√2 - 3√3 both numerator and denominator.

 \(\frac{2\sqrt3-\sqrt5}{2\sqrt2+3\sqrt3}\)

\(= \frac{(2\sqrt3-\sqrt5)(2\sqrt2-3\sqrt3)}{(2\sqrt2+3\sqrt3)(2\sqrt2-3\sqrt3)}\)

\(= \frac{(2\sqrt3-\sqrt5)(2\sqrt2-3\sqrt3)}{8-27}\)

\(=\frac{(4\sqrt6-2\sqrt10)-(18+3\sqrt15)}{-19}\)

\(= \frac{(18-4\sqrt6+2\sqrt10-3\sqrt15)}{19}\)

Given, \(\frac{\sqrt3-1}{\sqrt3+1}\) = x + y√3

= \(\frac{\sqrt3-1}{\sqrt3+1}\) x \(\frac{\sqrt3-1}{\sqrt3-1}\) = \(\frac{(\sqrt3-1)(\sqrt3-1)}{3-1}\)= \(\frac{4-2\sqrt3}{2}\) = 2 - √3

So, x = 2, y = -1

\(\sqrt{3+2\sqrt2}\)

\(= \sqrt{2+1+2\sqrt2}\)

\(= \sqrt{(\sqrt2)^2 + (1)^2 + 2 \times \sqrt2 \times 1}\)

^{\(= \sqrt{(\sqrt2+1)^2}\)}

\(= \sqrt2+1\)

(2 +√3)(2 -√3)

= (2)² - (√3)² [(a + b)(a - b) = a² – b²]

= 4 – 3 = 1.

\(\sqrt[5]{6}\) × \(\sqrt[5]{6}\) = (6)^{\(\frac{1}{5}\)} × (6)^{\(\frac{1}{5}\)} = (36)^{\(\frac{1}{5}\)}

= \(\sqrt[5]{36}\)

√10 × √15 =(√5×√2) × (√5×√3)

= 5 (√6)

(2 + √3) (2 – √3)

= (2)² – (√3)² [(a + b)(a – b) = a² – b²]

= 4 – 3

= 1

Given: a = √2 + 1

\(\frac{1}{a} = \frac{1}{(\sqrt2+1)}\)

\(= \frac{1}{(\sqrt2+1)}\) x \(\frac{(\sqrt2-1)}{(\sqrt2-1)}\)

\(= \frac{(\sqrt2-1)}{((\sqrt2)^2 - (1)^2)}\)

\(= \frac{(\sqrt2-1)}{1}\)

\(= \sqrt2-1\)

Now,

\(a - \frac{1}{a}\) = (√2 + 1) – (√2 – 1) 

= √2 + 1 - √2 + 1

= 2

Given: x = 2 + √3

\(\frac{1}{a} = \frac{1}{2+\sqrt3}\)

\(= \frac{1}{2+\sqrt3}\) x \(\frac{2-\sqrt3}{2-\sqrt3}\)

\(= \frac{(2-\sqrt3)}{((2)^2 - (\sqrt3)^2)}\)

\(= \frac{(2-\sqrt3)}{(4-3)}\)

\(= (2-\sqrt3)\)

Now,

\(x + \frac{1}{x}\) = (2 + √3) + (2 – √3)

= 2 + √3 + 2 – √3

= 4

Rationalisation factor of 7 – 3√5 is 7 + 3√5.

Rationalisation factor of √5 – 2 is √5 + 2.

Rationalisation factor of 2 +√3 = \(\frac{1}{2}\) +√3 = 2 -√3

√(13 - a√10) =√8 +√5

Squaring both side,..

= 13 – a√10 = 8 + 5 + 2 ×√8 ×√5

= 13 – a√10 = 13 + 2√40

= - a√10 = 4√10

= a = - 4

\(\frac{1}{\sqrt9-\sqrt8}\)

= \(\frac{1}{\sqrt9-\sqrt8}\) × \(\frac{\sqrt9 + \sqrt8}{\sqrt9+\sqrt8}\)

= √9 + √8 = 3 + 2√2

\(\sqrt{3-2\sqrt2}\)

(try to break the terms in form of (a+b)² or (a – b )²)

√(√2)²+ 1² – 2 ×√2×1) = √(√2-1)² = √2 – 1.

(i) (4 + \(\sqrt7\))(3 + \(\sqrt2\))

= 4 x 3 + 4 x \(\sqrt2\) + \(\sqrt7\) x 3 + \(\sqrt7\) x \(\sqrt2\)

= 12 + 4\(\sqrt2\) + 3\(\sqrt7\) + \(\sqrt14\)

(ii) (3 + \(\sqrt3\))(5 - \(\sqrt2\))

= 3 x 5 + 3 x (-\(\sqrt2\)) + \(\sqrt3\) x 5 + \(\sqrt3\) x (-\(\sqrt2\))

= 15 - 3\(\sqrt2\) + 5\(\sqrt3\) - \(\sqrt3\) x 2

= 15 - 3\(\sqrt2\)  + 5\(\sqrt3\) - \(\sqrt6\)

(iii) (\(\sqrt5\) - 2)(\(\sqrt3\) - \(\sqrt5\))

= \(\sqrt5\) x \(\sqrt3\) + \(\sqrt5\) x (-\(\sqrt5\)) + (-2) x \(\sqrt3\) +(-2) x (-\(\sqrt5\))

= \(\sqrt5\) x 3 - \(\sqrt5\) x 5 - 2\(\sqrt3\) + 2\(\sqrt5\)

= \(\sqrt15\) - \(\sqrt{5^2}\) - 2\(\sqrt3\) + 2\(\sqrt5\)

= \(\sqrt15\) - 5 - 2\(\sqrt3\) + 2\(\sqrt5\)

\(\sqrt{3-2\sqrt2}\) 

= \(\sqrt{(\sqrt{2})^2+(1)^2}\) - \(2\times \sqrt{2}+1\) 

= \(\sqrt{(\sqrt2-1)^2}\) 

= √2 - 1.

(11 + √11) (11 – √11) [(a + b)(a - b) = a² – b² ]

= 11² – (√11)² 

= 121 – 11 

= 110