Question

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

Answer 1

$x_1neq x_2neq x_3, \ left{begin{array}{c}x_1+x_2+x_3=27,\2x_2=x_1+x_3,\(x_2^2)^2=x_1^2cdot 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.$
$left{begin{array}{c}x_1+9+x_3=27,\x_2=9,\9^2=|x_1|cdot|x_3|;end{array}right. left{begin{array}{c}x_1+x_3=18,\x_2=9,\|x_1|cdot|x_3|=81;end{array}right. left{begin{array}{c}x_3=18-x_1,\x_2=9,\|x_1|cdot|18-x_1|=81;end{array}right.$
$left{begin{array}{c} left [ {{x_1(18-x_1)=81,} atop {-x_1(18-x_1)=81;}} right. \x_2=9,\x_3=18-x_1;end{array}right. left{begin{array}{c} left [ {{x_1^2-18x_1+81=0,} atop {x_1^2-18x_1-81=0;}} right. \x_2=9,\x_3=18-x_1;end{array}right. \ x_1^2-18x_1+81=0, \ (x_1-9)^2=0, \ x_1-9=0, \ x_1=9; \ x_1=x_2; \ x_1^2-18x_1-81=0, \ D_1=9^2+81=2cdot81, \ x_{11}=9-9sqrt{2}, x_{12}=9+9sqrt{2};$
$left{begin{array}{c} left [ {{x_1=9-9sqrt{2},} atop {x_1=9+9sqrt{2};}} right. \x_2=9,\x_3=18-x_1;end{array}right. left [ {{left{begin{array}{c} x_1=9-9sqrt{2}, \x_2=9,\x_3=9+9sqrt{2};end{array}right.} atop {left{begin{array}{c} x_1=9+9sqrt{2}, \x_2=9,\x_3=9-9sqrt{2}.end{array}right.}} right.$