Simplify- 111-2-111-4

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Byju's Answer

Standard VII

Mathematics

Power Rule of Exponents

Simplify: 111...

Question

Simplify: $\frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}}$

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Solution

Simplification.
BODMAS is the order of operations used to simplify the expressions
B - Brackets, O - Order of powers, D - Division, M - Multiplication, A - Addition, and S - Subtraction make up the abbreviation.
Simplify:
$\frac{11^{1 / 2}}{11^{1 / 4}} = 11^{(1 / 2) - (1 / 4)} [∵ a^{m} \times a^{- n} = a^{m - n}] = 11^{(1 / 4)}$
Therefore, the simplified value of the expression is $11^{1 / 2} / 11^{1 / 4} = 11^{1 / 4}$

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

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

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

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

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.

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Simplify: 11121114

Simplify: $\frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}}$