Question 180viSimplify -3-22 - 52-3- t-32-3-25 - 53-2- t-43

(3^ 2)^2 × (5^2)^ 3× (t^ 3)^2/(3^ 2)^5 × (5^3)^ 2× (t^ 4)^3 = (3)^ 4× (5)^ 6× (t)^ 6/(3)^ 10× (5)^ 6× (t)^ 12 [∵ (a^m)^n = (a)^mn] = (3)^ 4× (3)^10× (5)^ 6× ...

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Byju's Answer

Standard VII

Mathematics

Division of Exponents Having the Same Base

Question 180v...

Question

Question 180(vi)

Simplify :
$\frac{(3^{- 2})^{2} \times (5^{2})^{- 3} \times (t^{- 3})^{2}}{(3^{- 2})^{5} \times (5^{3})^{(- 2)} \times (t^{- 4})^{3}}$

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Solution

$\frac{(3^{- 2})^{2} \times (5^{2})^{- 3} \times (t^{- 3})^{2}}{(3^{- 2})^{5} \times (5^{3})^{- 2} \times (t^{- 4})^{3}} = \frac{(3)^{- 4} \times (5)^{- 6} \times (t)^{- 6}}{(3)^{- 10} \times (5)^{- 6} \times (t)^{- 12}} [∵ (a^{m})^{n} = (a)^{m n}] = (3)^{- 4} \times (3)^{10} \times (5)^{- 6} \times (5)^{6} \times (t)^{- 6} \times (t)^{12} = (3)^{- 4 + 10} \times (5)^{- 6 + 6} \times (t)^{- 6 + 12} [∵ a^{- m} = \frac{1}{a^{m}}] = (3)^{6} \times 5^{0} \times (t)^{6} = (3 t)^{6} [∵ a^{0} = 1]$

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Question 180(vi)

Simplify :
$\frac{(3^{- 2})^{2} \times (5^{2})^{- 3} \times (t^{- 3})^{2}}{(3^{- 2})^{5} \times (5^{3})^{(- 2)} \times (t^{- 4})^{3}}$

Q. Evaluate

\frac{{(3^{- 2})}^{2} \times {(5^{2})}^{- 3} \times {(t^{- 3})}^{1}}{{(3^{- 2})}^{5} \times {(5^{3})}^{- 2} \times {(t^{- 4})}^{3}}

Q. Evaluate:

\frac{(3 \frac{2}{3})^{2} - (2 \frac{1}{2})^{2}}{(4 \frac{3}{2})^{2} - (3 \frac{1}{3})^{2}}

\div

\frac{3 \frac{2}{3} - 2 \frac{1}{2}}{4 \frac{3}{2} - 3 \frac{1}{3}}

Q. Simplify:

\frac{{(3 \frac{2}{3})}^{2} - {(2 \frac{1}{2})}^{2}}{{(4 \frac{3}{4})}^{2} - {(3 \frac{1}{3})}^{2}} \div \frac{3 \frac{2}{3} - 2 \frac{1}{2}}{4 \frac{3}{4} - 3 \frac{1}{3}}

Find $(3^{3} - 2^{3}) + (5^{3} - 4^{3}) + (7^{3} - 6^{3}) + . . . .$ to 10 terms.

Show that $\frac{1 \times 2^{2} + 2 \times 3^{2} + . . . . + n \times (n + 1)^{2}}{1^{2} \times 2 + 2^{2} \times 3 + . . . . + n^{2} \times (n + 1)} = \frac{3 n + 5}{3 n + 1}$

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Question 180(vi) Simplify : (3−2)2×(52)−3×(t−3)2(3−2)5×(53)(−2)×(t−4)3

(3−2)2×(52)−3×(t−3)2(3−2)5×(53)−2×(t−4)3=(3)−4×(5)−6×(t)−6(3)−10×(5)−6×(t)−12 [∵(am)n=(a)mn]=(3)−4×(3)10×(5)−6×(5)6×(t)−6×(t)12=(3)−4+10×(5)−6+6×(t)−6+12 [∵a−m=1am]=(3)6×50×(t)6=(3t)6 [∵a0=1]

Question 180(vi)

Simplify :
$\frac{(3^{- 2})^{2} \times (5^{2})^{- 3} \times (t^{- 3})^{2}}{(3^{- 2})^{5} \times (5^{3})^{(- 2)} \times (t^{- 4})^{3}}$