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shuriki83
@shuriki83
June 2022
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(x+2)4+5(x+2)2-36=0
после скобок написаны степени
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pentV
(x+2)²=t
t²+5t-36=0
D=25+4*36=169=13²
t_1=(-5+13)/2=4; t_2=(-5-13)/2=-9
(x+2)²=4; x²+4x+4-4=0; x(x+4)=0; x=0 x=-4
(x+2)²=-9 корней нет
Ответ: x=0, x=-4
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Answers & Comments
t²+5t-36=0
D=25+4*36=169=13²
t_1=(-5+13)/2=4; t_2=(-5-13)/2=-9
(x+2)²=4; x²+4x+4-4=0; x(x+4)=0; x=0 x=-4
(x+2)²=-9 корней нет
Ответ: x=0, x=-4