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УСЛОВИЕ:
3 числа сост. арифметическую прогрессию. их сумма=27
а квадраты этих чисел сост. гелметрическую прогрессию. найти числа.


РЕШЕНИЕ:

$$ x_1\neq x_2\neq x_3, \\ \left\{\begin{array}{c}x_1+x_2+x_3=27,\\2x_2=x_1+x_3,\\(x_2^2)^2=x_1^2\cdot x_3^2;\end{array}\right. \left\{\begin{array}{c}x_1+x_2+x_3=27,\\-x_1+2x_2-x_3=0,\\x_2^2=|x_1|\cdot |x_3|;\end{array}\right. \left\{\begin{array}{c}x_1+x_2+x_3=27,\\3x_2=27,\\x_2^2=|x_1|\cdot |x_3|;\end{array}\right. $$
$$ \left\{\begin{array}{c}x_1+9+x_3=27,\\x_2=9,\\9^2=|x_1|\cdot|x_3|;\end{array}\right. \left\{\begin{array}{c}x_1+x_3=18,\\x_2=9,\\|x_1|\cdot|x_3|=81;\end{array}\right. \left\{\begin{array}{c}x_3=18-x_1,\\x_2=9,\\|x_1|\cdot|18-x_1|=81;\end{array}\right. $$
$$ \left\{\begin{array}{c} \left [ {{x_1(18-x_1)=81,} \atop {-x_1(18-x_1)=81;}} \right. \\x_2=9,\\x_3=18-x_1;\end{array}\right. \left\{\begin{array}{c} \left [ {{x_1^2-18x_1+81=0,} \atop {x_1^2-18x_1-81=0;}} \right. \\x_2=9,\\x_3=18-x_1;\end{array}\right. \\ x_1^2-18x_1+81=0, \\ (x_1-9)^2=0, \\ x_1-9=0, \\ x_1=9; \\ x_1=x_2; \\ x_1^2-18x_1-81=0, \\ D_1=9^2+81=2\cdot81, \\ x_{11}=9-9\sqrt{2}, x_{12}=9+9\sqrt{2}; $$
$$ \left\{\begin{array}{c} \left [ {{x_1=9-9\sqrt{2},} \atop {x_1=9+9\sqrt{2};}} \right. \\x_2=9,\\x_3=18-x_1;\end{array}\right. \left [ {{\left\{\begin{array}{c} x_1=9-9\sqrt{2}, \\x_2=9,\\x_3=9+9\sqrt{2};\end{array}\right.} \atop {\left\{\begin{array}{c} x_1=9+9\sqrt{2}, \\x_2=9,\\x_3=9-9\sqrt{2}.\end{array}\right.}} \right. $$






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