Set a^3=10 then this simplifies easily to ((a-1)/3)^5. Using Pascal’s triangle gives (a-1)^5=a^5-5.a^4+10.a^3-10.a^2+5.a-1=99-45a.Dividing by 3^5 gives the final solution (11-5a)/27.
Beautiful. But Sir, if we consider (5x + 1) = m instead of 5x = m, it looks bit simpler. So that the LHS becomes [(5x +1)/5x] ^ (5x + 1) which is [m/(m-1)]^m. We can transform it as [(m-1)/m]^-m. This is equal to (1 - (1/m))^ -m). This also can be written as (1 + (1/(-m))^ -m). This is when compared to the RHS, we get -m = 9 or m = -9. That is 5x + 1 = -9. Solving for x we get x = -2. This is just an inspiration from your work, no new idea. Thank you.
0 = (a^2 - b^2) - (b - a) = (a + b)(a - b) + (a - b) = (a - b)(a + b + 1). Thus, 0 = a - b ==> b = a or, similarly, b = -a - 1. From the first equation, a^2 - a = 73 or a^2 + a + 1 = 73. Each quadratic can be solved, whether the solutions are real or complex. They can then be tested in the second equation to see if the *system* is satisfied. I haven't begun to watch, but seeing a thumbnail claiming only one person was able to solve this is ... Odd.
why don't you just substitute y with it in terms of x, then square everything as needed? After rearranging terms and squaring a few times, you get the quadratic and the values of x. Plug back into the equation to get the y values.
Nice solution but here is an alternative: Define Z(n) = x^n +1/(x^n) => Z(0)=2, Z(1)=x+1/x, these have nice recursion formula: Z(n+1)=Z(n)*Z(1)-Z(n-1) (also Z(2n)=Z(n)^2-2). From this Z(7)=Z(1) * (Z(1)^6-7*Z(1)^4+14*Z(1)^2-7) = 3*281 =843=> Z(1)=3; Use recursion from [Z(1)....Z(7)]=[3,7,18,47,123,322,843]. Hence Z(3)+Z(5) = 18+123 = 141
Let t = 1 + 1/x + 12/x^2. Then the given equation is rewritten, after dividing both sides by x, as 1 = t/[sqrt(t+2)]. Squaring, t^2-t-2=0 > t = 2,-1. If t=2, x^2-x-12=0 > x = -3,4. t = -1 yields complex values for x, which we discard. Thus, x=-3,4.
Another solution: first of all I have found that 3-2-sqrt 2=( sqrt2-1)^2.. therefore the denominator becomes sqrt2-1.. 1/(sqrt-1)= sqrt2 +1.. then we have the cubic root of 63(sqrt2+1)-8, or of 63sqrt2+55.. If we suppose that this cubic root is a+b sqrt2, we can cube this binomial obtaining the system a^3+6ab^2=55; 3a^2b+2b^3=63.. the first one can be written as a(a^2+6b^2)=55..,55 can be written as 5*11 or 55*1..the only chance is a^2+6b^2=55 and a=1.. if we substitute we obtain b=3 and this couple satisfies the second equation too.. so we have 1+3sqrt2
This is clearly an example where the solution has been worked 'backwards' 😆i mean, you already know the solution, and that has dictated all your steps & choices right from the beginning. Such problems are, in a way, almost always useless.
Hi ! I did it so (2^x)-(3^x)=((2^(2x))*((2^x)-(3^x))^(1/3).Now lets take t=(2^x)-(3^x).Now take to the power 3 and you get t^3=((2)^2x))*t.Now 0 is obvious solution.Now divide by t ,you get t^2=2^2x.Now evaluate with x and you get (3/2)^x=2 and at the end x=log2/(log3-log2).It is less stresa ful method 😉.Greetings from Slovenia.
