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Решите задачу:

$1)\ \ \frac{(10x^4y ^{-3})^4}{16x^8(25x^2y^{-2})^3}= \frac{10^4x^{16}y^{-12}}{16x^825^3x^6y^{-6}} = \frac{10^4x^2y^{-6}}{16*25^3} = \frac{(5*2)^4x^2y^{-6}}{2^4*5^6} = \\ \\ =\frac{5^42^4x^2y^{-6}}{2^4*5^6} = 5^{-2}x^2y^{-6}= \frac{x^2}{25y^6} \\ \\ 2) \sqrt{72} +(3- \sqrt{2} )^2- \sqrt[4]{81} = \\ \\ \sqrt{4*2*9} +9-6 \sqrt{2}+2 - \sqrt[4]{3^4} =6\sqrt{2} +9-6 \sqrt{2} +2-3= 8 \\ \\ 3)\ x^{3} -3 x^{2} -4x+12=0 \\ x^{2} (x-3)-4(x-3)=0 \\ \\ (x-3)(x^2-4)=0 \\ (x-3)(x-2)(x+2)=0 \\$
$x=3\\x=2\\x=-2 \\ OTBET:3;\ 2;\ -2$