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Question 1: Rationalize the Denominator

Concept Explanation:

Rationalize denominator ka matlab hai ke denominator se root (√) ko hatana hai. Iske liye hum conjugate use karte hain.

Conjugate Rule:

• (a + b)(a – b) = a² – b²

• Agar denominator (a + √b) hai toh conjugate (a - √b) se multiply karein

• Agar denominator (a - √b) hai toh conjugate (a + √b) se multiply karein

(i) 13 / (4 + √3)

Step 1: Denominator mein (4 + √3) hai

Step 2: Conjugate se multiply karein: (4 - √3)

= 13/(4+√3) × (4-√3)/(4-√3)
= 13(4-√3) / (4+√3)(4-√3)
= 13(4-√3) / (16 - 3)
= 13(4-√3) / 13
= 4 - √3

✅ Answer: 4 - √3

(ii) (√2 + √5) / √3

Step 1: Denominator mein √3 hai

Step 2: √3 se multiply karein (numerator aur denominator dono ko)

= (√2 + √5)/√3 × √3/√3
= √3(√2 + √5) / 3
= (√6 + √15) / 3

✅ Answer: (√6 + √15) / 3

(iii) (√2 – 1) / √5

Step 1: Denominator mein √5 hai

Step 2: √5 se multiply karein

= (√2 - 1)/√5 × √5/√5
= √5(√2 - 1) / 5
= (√10 - √5) / 5

✅ Answer: (√10 - √5) / 5

(iv) (6 – 4√2) / (6 + 4√2)

Step 1: Denominator mein (6 + 4√2) hai

Step 2: Conjugate (6 - 4√2) se multiply karein

= (6-4√2)/(6+4√2) × (6-4√2)/(6-4√2)
= (6-4√2)² / (6+4√2)(6-4√2)
= (36 - 48√2 + 32) / (36 - 32)
= (68 - 48√2) / 4
= 17 - 12√2

✅ Answer: 17 - 12√2

(v) (√3 – √2) / (√3 + √2)

Step 1: Denominator mein (√3 + √2) hai

Step 2: Conjugate (√3 - √2) se multiply karein

= (√3-√2)/(√3+√2) × (√3-√2)/(√3-√2)
= (√3-√2)² / (√3+√2)(√3-√2)
= (3 - 2√6 + 2) / (3 - 2)
= (5 - 2√6) / 1
= 5 - 2√6

✅ Answer: 5 - 2√6

(vi) 4√3 / (√7 + √5)

Step 1: Denominator mein (√7 + √5) hai

Step 2: Conjugate (√7 - √5) se multiply karein

= 4√3/(√7+√5) × (√7-√5)/(√7-√5)
= 4√3(√7-√5) / (√7+√5)(√7-√5)
= 4√3(√7-√5) / (7 - 5)
= 4√3(√7-√5) / 2
= 2√3(√7-√5)
= 2√21 - 2√15

✅ Answer: 2√21 - 2√15

Question 2: Simplify the Following

Concept Explanation:

• (a/b)⁻ⁿ = (b/a)ⁿ

• aᵐ × aⁿ = aᵐ⁺ⁿ

• aᵐ ÷ aⁿ = aᵐ⁻ⁿ

• (aᵐ)ⁿ = aᵐⁿ

• a⁰ = 1

(i) (81/16)⁻³/⁴

= (16/81)³/⁴
= (2⁴ / 3⁴)³/⁴
= (2/3)⁴ × ³/⁴
= (2/3)³
= 8/27

✅ Answer: 8/27

(ii) (3/4)⁻² ÷ (4/9)³ × 16/27

Step 1: (3/4)⁻² = (4/3)² = 16/9

Step 2: (4/9)³ = 64/729

Step 3: 16/9 ÷ 64/729 × 16/27

= 16/9 × 729/64 × 16/27
= (16 × 729 × 16) / (9 × 64 × 27)
= (16 × 9 × 16) / (9 × 64)  [729/27 = 27, 27/9 = 3, 729/27 = 27]
= (16 × 16) / 64
= 256/64
= 4

✅ Answer: 4

(iii) (0.027)⁻¹/³

0.027 = 27/1000 = 3³/10³ = (3/10)³

(0.027)⁻¹/³ = (3/10)³ × ⁻¹/³
= (3/10)⁻¹
= 10/3

✅ Answer: 10/3

(iv) √(x³⁴y²¹z³⁵ / y³⁴z⁷)

= √(x³⁴ × y²¹⁻³⁴ × z³⁵⁻⁷)
= √(x³⁴ × y⁻¹³ × z²⁸)
= (x³⁴ × y⁻¹³ × z²⁸)¹/²
= x¹⁷ × y⁻¹³/² × z¹⁴
= x¹⁷ z¹⁴ / y¹³/²
= x¹⁷ z¹⁴ / √(y¹³)
= x¹⁷ z¹⁴ / (y¹³/²)

✅ Answer: x¹⁷ z¹⁴ / y¹³/²

(v) [5(25)ⁿ⁺¹ – 25(5)²ⁿ] / [5(5)²ⁿ⁺² – (25)ⁿ⁺¹]

Step 1: 25 = 5²

= [5(5²)ⁿ⁺¹ - 25(5)²ⁿ] / [5(5)²ⁿ⁺² - (5²)ⁿ⁺¹]
= [5(5)²ⁿ⁺² - 5²(5)²ⁿ] / [5(5)²ⁿ⁺² - (5)²ⁿ⁺²]
= [5(5)²ⁿ⁺² - 25(5)²ⁿ] / [5(5)²ⁿ⁺² - 5²ⁿ⁺²]
= [5(5)²ⁿ⁺² - 25(5)²ⁿ] / [5(5)²ⁿ⁺² - (5)²ⁿ⁺²]

Step 2: Let 5²ⁿ = a, 5² = 25

= [5 × 25 × 5²ⁿ - 25 × 5²ⁿ] / [5 × 25 × 5²ⁿ - 25 × 5²ⁿ]
= [125a - 25a] / [125a - 25a]
= 100a / 100a
= 1

✅ Answer: 1

(vi) [(16)ⁿ⁺¹ + 20(4²ⁿ)] / [2⁵ × 8ⁿ⁺²]

Step 1: 16 = 2⁴, 4 = 2², 8 = 2³

= [(2⁴)ⁿ⁺¹ + 20(2²)²ⁿ] / [2⁵ × (2³)ⁿ⁺²]
= [2⁴ⁿ⁺⁴ + 20(2⁴ⁿ)] / [2⁵ × 2³ⁿ⁺⁶]
= [2⁴ⁿ(2⁴ + 20)] / [2³ⁿ⁺¹¹]
= [2⁴ⁿ(16 + 20)] / [2³ⁿ⁺¹¹]
= [2⁴ⁿ(36)] / [2³ⁿ⁺¹¹]
= 36 × 2⁴ⁿ⁻³ⁿ⁻¹¹
= 36 × 2ⁿ⁻¹¹

✅ Answer: 36 × 2ⁿ⁻¹¹

(vii) (64)²/³ ÷ (9)³/²

(64)²/³ = (4³)²/³ = 4² = 16
(9)³/² = (3²)³/² = 3³ = 27

16 ÷ 27 = 16/27

✅ Answer: 16/27

(viii) (3ⁿ × 9ⁿ⁺¹) / (3ⁿ⁻¹ × 9ⁿ⁻¹)

= [3ⁿ × (3²)ⁿ⁺¹] / [3ⁿ⁻¹ × (3²)ⁿ⁻¹]
= [3ⁿ × 3²ⁿ⁺²] / [3ⁿ⁻¹ × 3²ⁿ⁻²]
= 3³ⁿ⁺² / 3³ⁿ⁻³
= 3⁵
= 243

✅ Answer: 243

(ix) (5ⁿ⁺³ – 6.5ⁿ⁺¹) / (9 × 5ⁿ – 2³ × 5ⁿ)

= [5ⁿ⁺¹(5² - 6)] / [5ⁿ(9 - 8)]
= [5ⁿ⁺¹(25 - 6)] / [5ⁿ(1)]
= [5ⁿ⁺¹(19)] / 5ⁿ
= 5 × 19
= 95

✅ Answer: 95

Question 3: If x = 3 + √8

Concept Explanation:

• x + 1/x ka formula use karein

• (a + b)² = a² + 2ab + b²

• (a – b)² = a² – 2ab + b²

Given: x = 3 + √8 = 3 + 2√2

Find 1/x: Conjugate use karein

1/x = 1/(3+2√2)
= (3-2√2)/(9-8)
= 3 - 2√2

(i) x + 1/x

= (3+2√2) + (3-2√2)
= 6

✅ Answer: 6

(ii) x – 1/x

= (3+2√2) - (3-2√2)
= 4√2

✅ Answer: 4√2

(iii) x² + 1/x²

Formula: x² + 1/x² = (x + 1/x)² – 2

= (6)² - 2
= 36 - 2
= 34

✅ Answer: 34

(iv) x² – 1/x²

Formula: x² – 1/x² = (x – 1/x)(x + 1/x)

= (4√2)(6)
= 24√2

✅ Answer: 24√2

(v) x⁴ + 1/x⁴

Formula: x⁴ + 1/x⁴ = (x² + 1/x²)² – 2

= (34)² - 2
= 1156 - 2
= 1154

✅ Answer: 1154

(vi) (x – 1/x)²

= (4√2)²
= 32

✅ Answer: 32

Question 4: Find p and q

Given: (8 – 3√2) / (4 + 3√2) = p + q√2

Step 1: Rationalize denominator

= (8-3√2)/(4+3√2) × (4-3√2)/(4-3√2)
= (8-3√2)(4-3√2) / (16 - 18)
= (32 - 24√2 - 12√2 + 18) / (-2)
= (50 - 36√2) / (-2)
= -25 + 18√2

Step 2: Compare with p + q√2

p = -25
q = 18

✅ Answer: p = -25, q = 18

Question 5: Simplify the Following

(i) [(25)³/² × (243)³/⁵] / [(16)¹/⁴ × (8)¹/³]

25 = 5² → (25)³/² = (5²)³/² = 5³ = 125
243 = 3⁵ → (243)³/⁵ = (3⁵)³/⁵ = 3³ = 27
16 = 2⁴ → (16)¹/⁴ = (2⁴)¹/⁴ = 2
8 = 2³ → (8)¹/³ = (2³)¹/³ = 2

= (125 × 27) / (2 × 2)
= 3375 / 4

✅ Answer: 3375/4

(ii) 54 × √(27)²ˣ / [9ⁿ⁺¹ + 216(3²ˣ⁻¹)]

Step 1: √(27)²ˣ = √(3³)²ˣ = √3⁶ˣ = 3³ˣ

Step 2: 54 = 2 × 27 = 2 × 3³

= 54 × 3³ˣ / [9ⁿ⁺¹ + 216(3²ˣ⁻¹)]
= (2 × 3³) × 3³ˣ / [3²ⁿ⁺² + 6³ × 3²ˣ⁻¹]
= 2 × 3³ˣ⁺³ / [3²ⁿ⁺² + 6³ × 3²ˣ⁻¹]
= 2 × 3³ˣ⁺³ / [3²ⁿ⁺² + 216 × 3²ˣ⁻¹]

Note: Iska further simplification possible nahi hai kyunke x aur n alag variables hain.

(iii) √[(216)²/³ × (25)⁻¹/² / (0.04)⁻³/²]

216 = 6³ → (216)²/³ = (6³)²/³ = 6² = 36
25 = 5² → (25)⁻¹/² = (5²)⁻¹/² = 5⁻¹ = 1/5
0.04 = 4/100 = 1/25 = (1/5)² → (0.04)⁻³/² = (1/5)⁻³ = 5³ = 125

= √[(36 × 1/5) / 125]
= √[36 / (5 × 125)]
= √[36 / 625]
= 6/25

✅ Answer: 6/25

(iv) (a¹/³ + b²/³) × (a²/³ – a¹/³b²/³ + b⁴/³)

Step 1: Let a¹/³ = x, b²/³ = y

Step 2: Expression becomes: (x + y)(x² – xy + y²)

= x³ + y³

Step 3: Substitute back:

= (a¹/³)³ + (b²/³)³
= a + b²

✅ Answer: a + b²