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An Important Olympiad Mathematics! 

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Hello friends,
In this video we are going to solve a nice Olympiad mathematics (algebra) using laws of indices.All are requested to learn this mathematics and if you like this video how to solve this problem please like share comment and subscribe to my channel.
#maths
#algebra
#mathsolympiadproblems
#niceequation
#entranceexam
#pleasesubscribe
Thanks 🙏👍 for watching!

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21 июн 2024

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Комментарии : 10   
@JoydeepBhattacharjee-to7jr
@JoydeepBhattacharjee-to7jr 17 дней назад
Nice technique to solve step by step
@annabanerjee5164
@annabanerjee5164 17 дней назад
Nice solution, please verify.
@DebdasBandyopadhyay-yq5jg
@DebdasBandyopadhyay-yq5jg 17 дней назад
Verified, please see in comment box.
@tinsokunsamphea5065
@tinsokunsamphea5065 2 дня назад
x=2;y=3;z=4 x=2;y=4;z=3 x=3;y=2;z=4 x=3;y=4;z=2 x=4;y=2;z=3 x=4;y=3;z=2
@walterwen2975
@walterwen2975 10 дней назад
An Important Olympiad Mathematics: 4^x + 4^y + 4^z = 336; x < y < z = ? 4^x + 4^y + 4^z = 336 = (16)(21) = 4²(1 + 4 + 16) = 4² + 4³ + 4⁴ x = 2, y = 3, z = 4 Answer check: 4^x + 4^y + 4^z = 336; Confirmed as shown Final answer: x = 2, y = 3, z = 4
@DebdasBandyopadhyay-yq5jg
@DebdasBandyopadhyay-yq5jg 17 дней назад
Verify: 4^x+4^y+4^z=336 4²+4³+4⁴ =16+64+256 =336 Therefore, X=2,y=3,z=4 is correct answer.
@marcgriselhubert3915
@marcgriselhubert3915 3 дня назад
336 = 16.21 = 16.(16 + 4 + 1) = (4^2).(4^2 + 4^1 + 4^0) = 4^4 + 4^3 + 4^2, then a = 4, b = 3, c = 2 (or other order)
@DebdasBandyopadhyay-yq5jg
@DebdasBandyopadhyay-yq5jg 3 дня назад
Thanks 🙏👍 for your comments
@taniacsibi6879
@taniacsibi6879 21 час назад
Ec. Are 6 soluții prin permutarea soluției (2 ,3,4 ) PTR.. in enunț nu se specifica nimic despre x, y,z
@DebdasBandyopadhyay-yq5jg
@DebdasBandyopadhyay-yq5jg 21 час назад
Thanks for your valuable information.
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