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A Nice Algebra Problem | Math Olympiad | How to solve for X in this problem ? 

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2 мар 2024

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Комментарии : 12   
@is7728
@is7728 4 месяца назад
9:22 I believe that it's easier to divide both sides by 4 first. Then simply calculate 57 * 50 = 57 * 100 / 2 = 5700 / 2 = 2850
@user-cg5xv4zz2b
@user-cg5xv4zz2b 3 месяца назад
this is a little tough if you go through it with all algebra - and I salute the presenter for going through it. It was fun to see the terms canceling ( 1/5 - 1/5 = 0 for examnple ). I chose to do it myself with - after the third line - getting common denominator and working the math - a lot less lines - but kinda needed a little calculator help- x( 1/21 + 1/77 + 1/165 + 1/285 ) = 200 cd = 3 5 7 11 19 = 21945 x( 1/(3*7) + 1/(7*11) + 1/( 3*5*11 ) + 1/( 3*5*19 ) ) = 200 x/21945 * ( (5*11*19) + ( 3*5*19 ) + ( 7*19 ) + (7*11) ) = 200 x/21945 * ( 1045 + 285 + 133 + 77 ) = 200 x/21945* ( 1540 ) = 200 x = ( 200 * 21945 ) /1540 x = 2850
@is7728
@is7728 4 месяца назад
This is really a typical question
@user-ee7nw2rx9s
@user-ee7nw2rx9s 3 месяца назад
Великолепно
@bertkoerts3991
@bertkoerts3991 3 месяца назад
Brilliant! I really enjoy and collect your challenges. Do you invent them yourselves? 👍😊
@pietrangelods9047
@pietrangelods9047 3 месяца назад
Least common multiple makes It Simplier
@DungNguyenvan-bg4gd
@DungNguyenvan-bg4gd 2 месяца назад
Thay vo ngau
@user-yp6gz6cy2v
@user-yp6gz6cy2v 3 месяца назад
Можно проще решить
@adilsonfigueiredo5544
@adilsonfigueiredo5544 3 месяца назад
Mmc, muito mais simples
@rafaellopez8946
@rafaellopez8946 3 месяца назад
Why not take mcm and Apply to the fractions?
@AnnaliaDalcero-dc8fl
@AnnaliaDalcero-dc8fl 3 месяца назад
Si fa piu' semplicemente
@user-wm4zm1nt9x
@user-wm4zm1nt9x 3 месяца назад
Это идиотизм решать такие задачи если ответ уже есть.
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