Відео

I too solved in less than a minute X = 4 and Y = 1 this gives a value of 243

There is a faster/simpler method a=5 b=7 c=8 cb = 35 bc = 56 ca = 40 a = 35/b c = 56/b ca = 40 c * a = 40 56/b * 35/b = 40 1960/b^2 = 40 1960 = 40b^2 49 = b^2 b = 7 a = 35/b a = 35/7 a = 5 c = 56/b c = 56/7 c = 8

I literally solved that in 2 seconds. 😂

Pretty clever, I though it was some horrible decimal number when I tried it.

Kul

Check it ?

Thanks a lot sir. Very nice explanation❤

Two comments 1. You can take i in the first step itself, and the solution becomes shorter. 2. Only negative value satisfies the given equation. Positive value 9i/2 is not a solution of the given equation.

2^×)^3+2^×=5^3+5 =>(2^×-5)+{(2^×)^3-5^5)}=0 =>(2^×-5)+(2^×-5){(2^2×+5(2^×)+5^2)}=0 =>Either 2^×=5.... =>×=log@2 (5) Or (2^×)^2+5(2^×)+26=0.. 2^× ={-5+_ _/5^2-4(26)}/2 Which goes for a complex soln.

3^× +×=30 => 1ln3=(30-×)Ln3 3:46 .e^Ln.3(-×) =>30.ln3.=30-×)Ln3 e^ln3(-×).e^ln3(30) =>W{27ln3.e^ln3(27) 8:43 } =W{30ln3.e^ln3(30-×)} =>27ln3=(30-×)Ln3 =>×=30-27=3 Firstly.....Lambert W-fnction.....W(×.e^×) 10:00 =× Secondly....3^30=( 10:00 10:00 =27ln3.e^ln3(27) =>×=30-27=3=

b

72/100