Question

prove that : 3^40 +3^39 +3^38 / 3^11 +3^40 +3^39 =13/33
${3}^{40} + {3}^{39} + {3}^{38} \div {3}^{41} + {3}^{40} - {3}^{39} = 13 \div 33$
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Answer 1

Steps to prove:

→  From the sum in the numerator and the denominator each, 3³⁸ has to be taken as common.

→  3³⁸ would be removed from the numerator and the denominator, thus there remains 3² + 3¹ + 3⁰ at the numerator and 3³ + 3² - 3¹ at the denominator.

→  Giving values 3³ = 27,  3² = 9,  3¹ = 3,  3⁰ = 1,  we get the RHS 13/33.

Preview:

$\displaystyle \Longrightarrow\ \frac{3^{40}+3^{39}+3^{38}}{3^{41}+3^{40}-3^{39}} \\ \\ \\ \Longrightarrow\ \frac{3^{38+2}+3^{38+1}+3^{38}}{3^{38+3}+3^{38+2}-3^{38+1}} \\ \\ \\ \Longrightarrow\ \frac{3^{38} \times 3^2+3^{38} \times 3^1+3^{38} \times 3^0}{3^{38} \times 3^3+3^{38} \times 3^2-3^{38} \times 3^1} \\ \\ \\ \Longrightarrow\ \frac{3^{38}[3^2+3^1+3^0]}{3^{38}[3^3+3^2-3^1]} \\ \\ \\ \Longrightarrow\ \frac{9+3+1}{27+9-3} \\ \\ \\ \Longrightarrow\ \frac{13}{33}$

Thus the answer is 13/33.

Hence proved!!!

Answer 2

Answer:

13/33 is the answer mark me as brainliest