Question

सरल कीजिए। (i) $\frac {25 \times t^{-4}}{5^{-3} \times 10 \times t^{-8}}\ (t \neq 0)$ (ii) $\frac {3^{-5} \times 10^{-5} \times 125}{5^{-7} \times 6^{-5}}$

Answer 1

Answer with Explanation:

(i) (25 × t⁻⁴)/(5⁻³ ×  10 × t⁻⁸)  

= (5² × t⁻⁴)/(5⁻³ ×  2¹ × 5¹ × t⁻⁸)        

[a^m ÷ a^n = a^m-n]        

= (5² ×  t⁻⁴ ⁻ ⁽⁻⁸⁾)/(5⁻³ ⁺ ¹  × 2¹)

= (5² ×  t⁻⁴ ⁺ ⁸)/(5⁻² × 2¹)

= (5²⁻ ⁽⁻²⁾ × t⁴)/2                            

[a^m ÷ a^n = a^m-n]        

= (5²⁺² × t⁴)/2

= (5⁴ × t⁴)/2

= 625 × t⁴/2                                      

 

(ii) (3⁻⁵ × 10⁻⁵ × 125)/(5⁻⁷ ×6⁻⁵)  

= {3⁻⁵ × (2 × 5)⁻⁵ × 5³}/(5⁻⁷  × 6⁻⁵)

= {3⁻⁵ × 2⁻⁵× 5⁻⁵ × 5³}/{5⁻⁷ × (2× 3)⁻⁵}

= {3⁻⁵ × 2⁻⁵ × 5⁻⁵ × 5³}/{5⁻⁷  × 2⁻⁵ × 3⁻⁵}

= 3⁻⁵ ⁻ ⁽⁻⁵⁾ × 2⁻⁵ ⁻ ⁽⁻⁵⁾ × 5⁻⁵ ⁺³ ⁻ ⁽⁻⁷⁾                          

[a^m ÷ a^n = a^m-n]        

= 3⁻⁵ ⁺ ⁵ × 2⁻⁵ ⁺⁵ × 5⁻⁵  ⁺³ ⁺⁷

= 3⁰× 2⁰ × 5⁻⁵ ⁺ ¹⁰

= 1 × 1 × 5⁵                                                  

  [a⁰ = 1]

= 3125

 आशा है कि यह उत्तर आपकी अवश्य मदद करेगा।।।।  

इस पाठ से संबंधित कुछ और प्रश्न :

मान ज्ञात कीजिए : (i) $\frac {8^{-1} \times 5^3}{2^{-4}}$ (ii) $(5^{-1} \times 2^{-1}) \times 6^{-1}$

https://brainly.in/question/10768626

मान ज्ञात कीजिए : (i) $(3^0 + 4^{-1}) \times 2^2$ (ii) $(2^{-1} \times 4^{-1}) \div 2^{-2}$ (iii) $(\frac{1}{2})^{-2} + (\frac{1}{3})^{-2} +(\frac{1}{4})^{-2}$ (iv) $(3^{-1} + 4^{-1} + 5^{-1})^0$ (v) $\{(\frac {-2}{3})^{-2}\}^2$

https://brainly.in/question/10768629

Answer 2

$\huge{\boxed{\red{\boxed{\mathfrak{\pink{Solution}}}}}}$

(i) (25 × t⁻⁴)/(5⁻³ × 10 × t⁻⁸)

= (5² × t⁻⁴)/(5⁻³ × 2¹ × 5¹ × t⁻⁸)

[a^m ÷ a^n = a^m-n]

= (5² × t⁻⁴ ⁻ ⁽⁻⁸⁾)/(5⁻³ ⁺ ¹ × 2¹)

= (5² × t⁻⁴ ⁺ ⁸)/(5⁻² × 2¹)

= (5²⁻ ⁽⁻²⁾ × t⁴)/2

[a^m ÷ a^n = a^m-n]

= (5²⁺² × t⁴)/2

= (5⁴ × t⁴)/2

= 625 × t⁴/2

(ii) (3⁻⁵ × 10⁻⁵ × 125)/(5⁻⁷ ×6⁻⁵)

= {3⁻⁵ × (2 × 5)⁻⁵ × 5³}/(5⁻⁷ × 6⁻⁵)

= {3⁻⁵ × 2⁻⁵× 5⁻⁵ × 5³}/{5⁻⁷ × (2× 3)⁻⁵}

= {3⁻⁵ × 2⁻⁵ × 5⁻⁵ × 5³}/{5⁻⁷ × 2⁻⁵ × 3⁻⁵}

= 3⁻⁵ ⁻ ⁽⁻⁵⁾ × 2⁻⁵ ⁻ ⁽⁻⁵⁾ × 5⁻⁵ ⁺³ ⁻ ⁽⁻⁷⁾

[a^m ÷ a^n = a^m-n]

= 3⁻⁵ ⁺ ⁵ × 2⁻⁵ ⁺⁵ × 5⁻⁵ ⁺³ ⁺⁷

= 3⁰× 2⁰ × 5⁻⁵ ⁺ ¹⁰

= 1 × 1 × 5⁵

[a⁰ = 1]

= 3125