Question

given ,√2=1.414and √6 =2.449,find the value of 1/√3-√2-1 to 3 places of decimal.

Answer 1

√6=2.449
√3×√2=2.449
√3=2.449/1.4/4 =1.731
1/1.731-1.414-1

=1/-0.683
=-1/0.683=
=-1.464

Answer 2

Answer:

$\frac{1}{\sqrt{3}-\sqrt{2}-1}=-1.464$

Step-by-step explanation:

Given : $\sqrt{2}=1.414$ and $\sqrt{6}=2.449$

To find : The value of $\frac{1}{\sqrt{3}-\sqrt{2}-1}$

Solution : First we find the value of $\sqrt{3}$

$\sqrt{6}=2.449$

$\sqrt{2}\times\sqrt{3}=2.449$

$1.414\times\sqrt{3}=2.449$

$\sqrt{3}=\frac{2.449}{1.414}$

$\sqrt{3}=1.731$

Now, put all the values in the expression

$\frac{1}{\sqrt{3}-\sqrt{2}-1}$

$=\frac{1}{1.731-1.414-1}$

$=\frac{1}{-0.683}$

$=-1.4641$

Up to 3 decimal place = -1.464