MATHS IN READING ROOM! এই নামের চ্যানেল থেকে আমরা অঙ্কের সমাধান জানব। অঙ্কের সমাধান করতে এই চ্যানেলটি ফলো করার অনুরোধ করছি। অঙ্কের ভীতি দূর করতে সাহায্য করবে এই চ্যানেলটি। সকল দর্শক, ছাত্র ছাত্রীদের উপকার হলে আমার শ্রম সার্থক হবে। সকলকে ধন্যবাদ জানাই। আশা করি সাবস্ক্রাইব করে রাখবেন। CHANNEL NAME IS:-- MATHS IN READING ROOM! FROM THIS CHANNEL We will learn mathematics. All are requested to see the videos. I hope this helps them to solve math problem. All are requested to subscribe my channel. Thanks for watching.
All maths lovers are requested to watch this mathematics.This mathematics is very important for Olympiad Question also competitive exams. We should learn this mathematics.
(9^2001-9^1999)/(3^4000-3^3996)= (9^2001-9^1999)/((3^(2*2000)-3^(2*1998))= | 4000=2*2000 and 3996=2*1998 (9^2001--9^1999)/((3^2)^2000-(3^2)^1998)= | a^(m*n)=(a^m)^n (9^2001--9^1999)/(9^2000-9^1998)= | 3^2=9 (9^(2+1999)-9^1999)/(9^(2+1998)-9^1998)= | a^(m+n)=a^m*a^n (9^2*9^1999--1*9^1999)/(9^2*9^1998-1*9^1998) | 9^1999 is common in enumerator and 9^1998 is common in denominator 9^1999*(9^2-1)/(9^1998*(9^2-1)= | (9^2--1) is common in enumerator and denominator 9^1999/9^1998= | a^m/a^n=a^(m-n) 9^(1999-1998)= 9^1= 9 I think, it is easier to take 9^1998 out of the sum in the enumerator and to take out 9^1996 out of the sum in the denominator If you do that, you would not get fractions divided b fractions like in the video.
(First: x <>0 mod Pi/2 We also notice that tan(Pi/8) = sqrt(2) - 1 and that cotan(Pi/8) = sqrt(2) +1) We have (1/tan(x)) -tan(x) = 2 or (tan(x)^2 +2.tan(x) -1 = 0 Then tan(x) = sqrt(2) - 1 or tan(x) = -(sqrt(2) +1) *If tan(x) = sqrt(2) - 1 = tan(Pi/8) then x = Pi/8 mod Pi *If tan(x) = -(sqrt(2) - 1) = -cotan(Pi/8) = -tan(Pi/2 - Pi/8) = tan(-3.Pi/8) then x = -3.Pi/8 mod Pi *We can sum up: x = Pi/8 mod Pi/2 (as -3.Pi/8 = Pi/8 - Pi/2)
Also my method, but I did [3^3,998(3^4 - 1)]/[3^3,996(3^4 - 1)]. Same idea, different numbers. Just as simple for us, and vastly more simple than the video.
@@DebdasBandyopadhyay-yq5jg I agree, Debdas, but why didn't you not take out the 9^1999 from numerator and 3^3996 from denominator, and in 2 steps the answer would be 9.
Thanks for your valuable information. Sometimes we get a root that doesn't satisfy the original equation.This type of roots are called extraneous root. So we should verify the the answer.Here 1/5 is a extraneous root! Thank you.
