Math, asked by Al8sha, 1 year ago

hey...
....
.....
plz help

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Answers

Answered by DaIncredible
1
Heya dear,
Here is the answer you were looking for:

Given,
√2 = 1.414

 \sqrt{ \frac{ \sqrt{2  -  1} }{ \sqrt{2 + 1} } }  \\  \\  =  \sqrt{ \frac{ \sqrt{2}  -  \sqrt{1} }{ \sqrt{2} +  \sqrt{1}  } }  \\

On rationalizing the denominator we get,

 =  \sqrt{ \frac{ \sqrt{2}  -  \sqrt{1} }{ \sqrt{2} +  \sqrt{1}  } \times  \frac{ \sqrt{2} -  \sqrt{1}  }{ \sqrt{2}   -  \sqrt{1} }  }  \\

Using the identity:

 {(x - y)}^{2}  =  {x}^{2}  +  {y}^{2}  - 2xy \\ (x + y)(x - y) =  {x}^{2}  -  {y}^{2}

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


We can write it as

 \sqrt{ { (\sqrt{2} )}^{2} +  {( \sqrt{1}) }^{2}  - 2( \sqrt{2})( \sqrt{1})   }  \\  \\  =   \sqrt{ {( \sqrt{2}  +  \sqrt{1} )}^{2} }  \\  \\  =  \sqrt{2}  +  \sqrt{1}  \\  \\  =  \sqrt{2}  +  {( \sqrt{1} )}^{2}  \\  \\  =  \sqrt{2}  + 1 \\  \\  = 1.414 + 1 \\  \\  = 2.414 \\


Hope this helps!!

If you have any doubt regarding to my answer, feel free to ask in the comment section or inbox me if needed.

@Mahak24

Thanks...
☺☺

Al8sha: thank you so much
DaIncredible: my pleasure... glad to help ^_^
Al8sha: Can you explain the next 4 steps after whole root over 3-2root 2
DaIncredible: we can also write it as √[(√2)^2 + (√1)^2 - 2(√2)(√1)
DaIncredible: same thing
Al8sha: okk, but after this step it will be whole root over root 2- root 1 whole square...
DaIncredible: yep
DaIncredible: we know that √(a)^2 = a
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