Advertisements
Advertisements
प्रश्न
If \[\sqrt{2} = 1 . 4142\] then \[\sqrt{\frac{\sqrt{2} - 1}{\sqrt{2} + 1}}\] is equal to
विकल्प
0.1718
5.8282
0.4142
2.4142
Advertisements
उत्तर
0.4142
Explanation:
Given that `sqrt2= 1.4142`, we need to find the value of `sqrt((sqrt2-1)/(sqrt2+1))`
We can rationalize the denominator of the given expression. We know that rationalization factor for `sqrt2+1` is`sqrt2-1`. We will multiply numerator and denominator of the given expression `sqrt((sqrt2-1)/(sqrt2+1))`by`sqrt2-1`, to get
`sqrt((sqrt2-1)/(sqrt2+1)) = sqrt((sqrt2-1)/(sqrt2+1)xxsqrt((sqrt2-1)/(sqrt2-1)))`
` = sqrt((sqrt2-1)^2/((sqrt2)^2-1))`
` = sqrt((sqrt2-1)^2)/(sqrt((sqrt2)^2-1))`
\[\sqrt{\frac{\sqrt{2} - 1}{\sqrt{2} + 1}} = \frac{\sqrt{2} - 1}{1}\]
Putting the value of `sqrt2`, we get
`sqrt2 -1 = 4.4142 - 1`
`= 0.4142`
APPEARS IN
संबंधित प्रश्न
Simplify:-
`2^(2/3). 2^(1/5)`
Simplify the following
`((x^2y^2)/(a^2b^3))^n`
Simplify the following
`(a^(3n-9))^6/(a^(2n-4))`
Show that:
`(3^a/3^b)^(a+b)(3^b/3^c)^(b+c)(3^c/3^a)^(c+a)=1`
Solve the following equation:
`3^(x-1)xx5^(2y-3)=225`
If 1176 = `2^axx3^bxx7^c,` find the values of a, b and c. Hence, compute the value of `2^axx3^bxx7^-c` as a fraction.
If a, b, c are positive real numbers, then \[\sqrt{a^{- 1} b} \times \sqrt{b^{- 1} c} \times \sqrt{c^{- 1} a}\] is equal to
If 102y = 25, then 10-y equals
If \[\sqrt{2^n} = 1024,\] then \[{3^2}^\left( \frac{n}{4} - 4 \right) =\]
If \[\sqrt{13 - a\sqrt{10}} = \sqrt{8} + \sqrt{5}, \text { then a } =\]
