Question

If a = √5 +1/√5 -1 and b= √5-1/√5+1, then find a^2+ab+b^2/a^2-ab+b^2.​

Answer 1

Step-by-step explanation:

Solution

a² + ab + b²

a²ab +6²

(a² +6²) + ab

(a² + 6²) -ab

(a + b)² - 2ab + ab

(a + b)²-2ab-ab

[a² + b² = (a + b)² - 2ab]

(a + b)² - ab

(a + b)²-3ab

(1)

√5+1

√5-1

a =

b =

√5 1

√5+1

ab =

√5 +1

√5 -1

√5+1

X

= 1√5 +1 √5-1

√5-1

√5+1

ab =

√5 +1 √5-1

+

a+b=

√5-1 √5 +1

(√5 + 1)² + (√5 - 1)²

(√5 + 1)(√5 - 1)

2[(√5)² + (1)²]

(√5)²-(1)²

(a + b)² + (a - b)² = 2(a² +6²)

(a + b) + (a - b) = a² - b²

2(5+1)

5-1

2 × 6

4

= 3

... From (1)→

a² + ab +6²

a²ab + b²

(3)²-.. From (1)

a² + ab + b²

a²ab + b²

(3)² - 1

(3)² - 3 x 1

9-1

9-3

8

4

3

Hence, value of

a² + ab + b²

a²ab + b²

4

3

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