If $$x=(9+4\sqrt{5})^{48} = [x] +f$$, where [x] is defined as integral part of x and f is a fraction, then x (

Question

If $$x=(9+4\sqrt{5})^{48} = [x] +f$$, where [x] is defined as integral part of x and f is a fraction, then x (1 - f) equals

Accepted Answer

1. It is given that x = ( 9 + 4 5 ) 48 x=(9+4\sqrt{5})^{48} x = ( 9 + 4 5 ​ ) 48 ... (1) Let us assume that y = ( 9 − 4 5 ) 48 y=(9-4\sqrt{5})^{48} y = ( 9 − 4 5 ​ ) 48 ... (2) We can see that 0 < y < 1. Also, x + y = 2( 48 C 0 ∗ ( 9 ) 48 48C0*(9)^{48} 48 C 0 ∗ ( 9 ) 48 + 48 C 2 ∗ ( 9 ) 46 ∗ ( 4

If $x=(9+4\sqrt{5})^{48} = [x] +f$ , where [x] is defined as integral part of x and f is a fraction, then x (1 - f) equals

If $x=(9+4\sqrt{5})^{48} = [x] +f$ , where [x] is defined as integral part of x and f is a fraction, then x (1 - f) equals

If x=(9+45)48=[x]+fx=(9+4\sqrt{5})^{48} = [x] +fx=(9+45​)48=[x]+f, where [x] is defined as integral part of x and f is a fraction, then x (1 - f) equals

If x=(9+45)48=[x]+fx=(9+4\sqrt{5})^{48} = [x] +fx=(9+45​)48=[x]+f, where [x] is defined as integral part of x and f is a fraction, then x (1 - f) equals

If $x=(9+4\sqrt{5})^{48} = [x] +f$ , where [x] is defined as integral part of x and f is a fraction, then x (1 - f) equals

If $x=(9+4\sqrt{5})^{48} = [x] +f$ , where [x] is defined as integral part of x and f is a fraction, then x (1 - f) equals