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Решите задачу:

$\left \{ {{x(1+y)=66} \atop {x^2y=128}} \right. \\ \\ x(1+ \frac{128}{x^2} )=66 \\ \\ \\ x+ \frac{128}{x} =66 \\ \\ x^2-66x+128=0 \\$
$\left \{ {{x(1+y)=66} \atop {x^2y=128}} \right. \\ \\ x(1+ \frac{128}{x^2} )=66 \\ \\ \\ x+ \frac{128}{x} =66 \\ \\ x^2-66x+128=0 \\ \\ D=4356-512=3844=62^2 \\ \\ x_{1} =(66-62)/2=2; y_{1} =128/4=32 \\ \\ x_{2} =(66+62)/2=64; y_{2} =128/4096=1/32$