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Комментарии : 13   
@georgesbv1
@georgesbv1 7 дней назад
Also possible to calculate using factorization. We found prime factors and we get 11 as largest prime . Then we arrange so that 11 × 10 × 9 x 8= 7920
@tunneloflight
@tunneloflight 7 дней назад
I did the same. Along with the other factors 2,2,2,2,3,3,5,11 -> 2*2*2=8, 3*3=9, 2*5=10, 11. For me it was faster to est. the fourth root by using long division (8 into 79 = 64 remainder 15, new divisor 16_ into 1520 = 9, so 89. 9 into 89 result 9 check remainder 8. new divisor 18_ into 800 = 4, so 9.4. Hence 9 and 10 as middle two numbers, 8 and 11 as the other two. But it was pretty close to an even race. The time doing the divisions to get the factors took longer than the root extractions. Then reassembling those took a moment. If I hadn't had a lot of practice doing square roots by hand, the factors would have been faster. Usually though, the first two digits of square roots is easy enough to do it my head without having to write it down. By the third digit it is too easy to make errors. In this case, two digits was more than enough.
@tunneloflight
@tunneloflight 7 дней назад
The result is the product of 4 consecutive integers. Estimate the 4th root of 7920 by long division method = about 9.4 ( sqrt of 7920^(1/2)= about 89; 89^(1/2) = about 9.4. So the two middle values must be 9 and 10, the others are 8 and 11. Check 8*9*10*11=7,920 -> Check. X+7 = 11 => x=4. And the LHS = 11! / 7! = 7,920 QED. Result: X = 4.
@MrGeorge1896
@MrGeorge1896 6 дней назад
That is the usual way to solve this kind of problems but you can't fill a 20+ min video with such a quick and easy solution 😅
@tunneloflight
@tunneloflight 6 дней назад
@@MrGeorge1896 Well, in that case... Convert the numbers to base 3 to increase the difficulty, and also convert the factorials into gamma functions.
@juergenilse3259
@juergenilse3259 6 дней назад
(x+7)!/(x+3)!=7920 (x+4)*(x+5)*(x+6)*(x+7)=7920 (x+4)*(x+5)*(x+6)*(x+7)=72*110=2*36*110=2*4*9*10*11=8*9*10*11 comparing the factors llead to x+4=8 and x+5=9 and x+6=10 and x+7=11 so x must be 4..
@RyanLewis-Johnson-wq6xs
@RyanLewis-Johnson-wq6xs 6 дней назад
(X+7)!/(X+3)!=7920 X=4 X=-5.5 ± 1.5Sqrt[39]i=(-11 ±3Sqrt[39]i)/2
@kksomdr
@kksomdr 6 дней назад
Let x^2+11x+29=t. (t-1)(t+1)=7920. t^2-1=7920. t^2=7921.....
@latharajesh8376
@latharajesh8376 16 часов назад
x=4
@rainerrosenberger9111
@rainerrosenberger9111 7 дней назад
Awfull awkward solution, previous comments Show how to solve it faster and better
@ManojkantSamal
@ManojkantSamal 6 дней назад
(X+7)!= x(x+1)(x+2)(x+4)(x+5)(x+6)(x+7) (X+3)!=x(x+1)(x+2)(x+3) (X+7)!/(x+3)!= (X+4)(x+5)(x+6)(x+7) So {(X+4)(x+7)}{(x+5)(x+6)} =(x^2+11x+28)(x^2+11x+30) Let X^2+11x+28=a So a(a+2)=a^2+2a As per question a^2+2a=7920 a^2+2a-7920=0 a^2+90a-88a-7920=0 a(a+90)-88(a+90)=0 (a+90)(a-88)=0 a=90 or 88 X^2+11x+28=88 X^2+11x-60=0 X^2+15x-4x-60=0 X(x+15)-4(x+15)=0 (X+15)(x-4)=0 X+15=0 or x-4=0 X=-15 or x=4 Another pair of values are ther while we put x=90
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