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I also used substitution. Let u = x - 4, so the equation will be u! = u + 3. As 3 must be divisible by u, only 1, 2 or 3 are possible solutions. So we get u = 3 and x = 3 + 4 = 7.
Let a=(x-4), a! =a+3, a=3 then, x=7👍
Let x-4=t Then, t!=t+3 > t=3 > x=7.
x-4 =uThen problem is reduced tou[(u-1)! -1] =3So u =1 or u=3u can't be 1 as 2nd term is -veu =3Then x = 7
Solved in 10’’ : 4, 5, 6 do not work, 7 works inequality is obvious starting from 8. X=7.
x= 7
(x ➖ 4)^2=( x^2 ➖ 16)= 14 2^7 (x ➖ 7x+2). (x ➖ 1)^2= (x^2 ➖ 1)= x^1 (x ➖ 1x+1) .
I also used substitution. Let u = x - 4, so the equation will be u! = u + 3. As 3 must be divisible by u, only 1, 2 or 3 are possible solutions. So we get u = 3 and x = 3 + 4 = 7.
Let a=(x-4), a! =a+3, a=3 then, x=7👍
Let x-4=t Then, t!=t+3 > t=3 > x=7.
x-4 =u
Then problem is reduced to
u[(u-1)! -1] =3
So u =1 or u=3
u can't be 1 as 2nd term is -ve
u =3
Then x = 7
Solved in 10’’ : 4, 5, 6 do not work, 7 works inequality is obvious starting from 8. X=7.
x= 7
(x ➖ 4)^2=( x^2 ➖ 16)= 14 2^7 (x ➖ 7x+2). (x ➖ 1)^2= (x^2 ➖ 1)= x^1 (x ➖ 1x+1) .