Question

если x-y=2√3  x+y=3√2 то разность x^4-y^4 равно

Answer 1

 
 $x-y=2sqrt{3}\ x+y=3sqrt{2}\ \ x^4-y^4=(x^2-y^2)(x^2+y^2)=(x-y)(x+y)(x^2+y^2)\\ x^4-y^4=(x-y)(x+y)((x+y)^2-2xy)\\ x^2-2xy+y^2=12\ x^2+2xy+y^2=18\\ xy=frac{3}{2}\\ x^4-y^4=2sqrt{3}*3sqrt{2}(18-3)=6sqrt{6} * 15 = 90*sqtr{6}$