Simplify 1/√(2)by rationalizing the denominator.[ 1/√(2)=1/√(2)×√(2)/√(2); =√(2)/2 ]Hence the simplified form of 1/√(2) is √(2)/2. ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard IX

Mathematics

Fractional Exponent

Simplify : 1/...

Question

Simplify : $\frac{1}{\sqrt{2}}$

Open in App

Solution

Simplify $\frac{1}{\sqrt{2}}$ by rationalizing the denominator.
$\begin{matrix} \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} \\ = \frac{\sqrt{2}}{2} \end{matrix}$
Hence the simplified form of $\frac{1}{\sqrt{2}}$ is $\frac{\sqrt{2}}{2}$ .

Suggest Corrections

Similar questions

${\frac{1}{(s e c^{2} θ - c o s^{2} θ)} + \frac{1}{(c o s e c^{2} θ - s i n^{2} θ)}} (s i n^{2} θ c o s^{2} θ) = \frac{1 - s i n^{2} θ c o s^{2} θ}{2 + s i n^{2} θ c o s^{2} θ}$

Q. If

α

β

γ

are the roots of

x^{3} + a x^{2} + b = 0

b \neq 0

then the determinant

Δ

, where

Δ = ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{1}{α} & \frac{1}{β} & \frac{1}{γ} \frac{1}{β} & \frac{1}{γ} & \frac{1}{α} \frac{1}{γ} & \frac{1}{α} & \frac{1}{β} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣

equals

Determine whether the following numbers are in proportion or not $:$

$\frac{1}{3}, \frac{1}{4}, \frac{1}{6}, \frac{1}{7}$

Q. Among

\frac{1}{12}

\frac{1}{15}

and

\frac{1}{17}

, is the greatest fraction

Q. Assertion :

Δ = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{(1 + a_{1} b_{1}) (1 + a_{1}^{2} b_{1}^{2} - a_{1} b_{1})}{1 + a_{1} b_{1}} & \frac{(1 + a_{1} b_{2}) (1 + a_{1}^{2} b_{2}^{2} - a_{1} b_{2})}{1 + a_{1} b_{2}} & \frac{(1 + a_{1} b_{3}) (1 + a_{1}^{2} b_{3}^{2} - a_{1} b_{3})}{1 + a_{1} b_{3}} \frac{(1 + a_{2} b_{1}) (1 + a_{2}^{2} b_{1}^{2} - a_{2} b_{1})}{1 + a_{2} b_{1}} & \frac{(1 + a_{2} b_{2}) (1 + a_{2}^{2} b_{2}^{2} - a_{2} b_{2})}{1 + a_{2} b_{2}} & \frac{(1 + a_{2} b_{3}) (1 + a_{2}^{2} b_{3}^{2} - a_{2} b_{3})}{1 + a_{2} b_{3}} \frac{(1 + a_{3} b_{1}) (1 + a_{3}^{2} b_{1}^{2} - a_{3} b_{1})}{1 + a_{3} b_{1}} & \frac{(1 + a_{3} b_{2}) (1 + a_{2}^{2} b_{2}^{2} - a_{3} b_{2})}{1 + a_{3} b_{2}} & \frac{(1 + a_{3} b_{3}) (1 + a_{3}^{2} b_{3}^{2} - a_{3} b_{3})}{1 + a_{3} b_{3}} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

Δ = 0

Reason:

Δ

can be written as product of two determinants.

Join BYJU'S Learning Program

Simplify : 12

Simplify 12by rationalizing the denominator.12=12×22=22Hence the simplified form of 12 is 22.

Simplify : $\frac{1}{\sqrt{2}}$

Simplify $\frac{1}{\sqrt{2}}$ by rationalizing the denominator.
$\begin{matrix} \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} \\ = \frac{\sqrt{2}}{2} \end{matrix}$
Hence the simplified form of $\frac{1}{\sqrt{2}}$ is $\frac{\sqrt{2}}{2}$ .