I think it is simpler to work this way: Move all terms in one side (x^x)²-x²(x^x)-(x^x)x+x³=0 Factorize (x^x)[(x^x)-x²]-x[(x^x)-x²]=0 [(x^x)-x²][(x^x)-x]=0 (x^x)=x² --> x^(x-2)=1 x=1 or x=2 (x^x)=x --> x=±1 Thus x={-1,1 2}
@williamspostoronnim9845
2 ай бұрын
Ничего себе примерчик!
@hookahsupplier.5155
8 ай бұрын
Amazing question, didn't solve it but had fun watching the solution!
@MATHUP869
8 ай бұрын
Thank you for your shareing Master
@SyberMath
8 ай бұрын
My pleasure
@scottleung9587
8 ай бұрын
I got 1 and 2 but didn't consider -1.
@leif1075
8 ай бұрын
Youbforgot zero too..also a valid solution..and maybe i in the complex world
@altosrule6387
8 ай бұрын
@@leif1075 Zero is not a valid solution. The limit as x goes to 0 of x^(2x) is 1, while the other terms all equal zero. I don't think i is a valid solution either.
@Mdsamim-go8rp
8 ай бұрын
Beautiful equation with Beautiful explanation ❤❤
@SyberMath
8 ай бұрын
Thanks a lot 😊❤️
@markener4316
8 ай бұрын
What would if theroaticlly possible a integrated matrix be?
Пікірлер: 15