B- pqrs x C- qr - ps 0-2 xD - ps - qr 2 x

The correct option is A 1Using law of exponents, a^m/b^m=(a/b)^m. p^xq^xr^xs^x=(pq)^x(rs)^x =(pqrs)^x =(psqr)^x =(ps)^x(qr)^x Also, (rqps)^x =(rq)^x(ps)^x. p^ ...

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Standard VIII

Mathematics

Dividing Terms with the Same Exponent

B. pqrs x C. ...

Question

$\frac{\frac{p^{x}}{q^{x}}}{\frac{r^{x}}{s^{x}}} \times {(\frac{r q}{p s})}^{x} =$ ___

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{(\frac{q r}{p s})}^{2 x}

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{(\frac{p s}{q r})}^{2 x}

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(p q r s)^{x}

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Solution

The correct option is A 1
Using law of exponents, $\frac{a^{m}}{b^{m}} = (\frac{a}{b})^{m} .$
$⟹ \frac{\frac{p^{x}}{q^{x}}}{\frac{r^{x}}{s^{x}}} = \frac{{(\frac{p}{q})}^{x}}{{(\frac{r}{s})}^{x}}$
$= {⎛ ⎜ ⎜ ⎜ ⎝ \frac{\frac{p}{q}}{\frac{r}{s}} ⎞ ⎟ ⎟ ⎟ ⎠}^{x}$
$= {(\frac{p s}{q r})}^{x}$
$= \frac{(p s)^{x}}{(q r)^{x}}$
Also, ${(\frac{r q}{p s})}^{x} = \frac{(r q)^{x}}{(p s)^{x}} .$

$⟹ \frac{\frac{p^{x}}{q^{x}}}{\frac{r^{x}}{s^{x}}} \times {(\frac{r q}{p s})}^{x} = \frac{(p s)^{x}}{(q r)^{x}} \times \frac{(r q)^{x}}{(p s)^{x}}$
$= 1$

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Similar questions

Question 9

In $Δ P Q R$ ,

a) PQ - QR > PR
b) PQ + QR < PR
c) PQ - QR < PR
d) PQ + PR < QR

Q. In

△ P Q R, ∠ P Q R = 90^{o}, Q S ⊥ P R

. If

P Q = a

Q R = b

R P = c

and

Q S = p

, show that

p c = a b

Q. In

△ P Q R,

P D ⊥ Q R

such that

D

lies on

Q R .

P Q = a, P R = b, Q D = c

and

D R = d,

then

Q. PQRS is a parallelogram in which

∠ Q R S

and

∠ P Q R

are in ratio 7 : 3. Find the sum of

∠ P Q R

and

∠ P R Q .

\frac{- 34 p^{3} q^{2} r}{17 p^{2} q}

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pxqxrxsx×(rqps)x= ___

The correct option is A 1Using law of exponents, ambm=(ab)m. ⟹pxqxrxsx=(pq)x(rs)x =⎛⎜ ⎜ ⎜⎝pqrs⎞⎟ ⎟ ⎟⎠x =(psqr)x =(ps)x(qr)x Also, (rqps)x=(rq)x(ps)x. ⟹pxqxrxsx×(rqps)x=(ps)x(qr)x×(rq)x(ps)x =1

$\frac{\frac{p^{x}}{q^{x}}}{\frac{r^{x}}{s^{x}}} \times {(\frac{r q}{p s})}^{x} =$ ___