Question

solve this 9th class ch 1 hard but Interesting question.
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Answer 1

Answer:

To get the solution of this first ,Let y = [{√ (√5+2) +√ (√5-2)} /√ (√5+1)]-√ (3-2√2)

Now, let x={√ (√5+2) +√ (√5-2)} /√ (√5+1)

then x^2=[√5+2+√5-2+2*√(5–4)]/(√5+1) = (2√5+2)/(√5+1) = 2

x^2= 2

so x={√ (√5+2) +√ (√5-2)} /√ (√5+1) = √2

So now question is y= x - √ (3-2√2)

i.e. y= √2 - √ (3-2√2)

now, let z= √ (3-2√2) = √ (2+1-2√2) = √[{ (√2 )²}+1-2*√2*1] = √[(√2 – 1)²] = √2 – 1

So y = √2- (√2 – 1 )

So the value of y = 1.