Question

$3(x^{2} + \frac{1}{x ^{2} } ) + 7(x + \frac{1}{x} ) = 4$
Нужно решить с помощью замены
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Answer 1

$3(x^2+\frac{1}{x^2} )+7(x+\frac{1}{x} )=4\\\\(x+\frac{1}{x} )^2=x^2+2+\frac{1}{x^2} \\\\x+\frac{1}{x}=a \\x^2+\frac{1}{x^2}=a^2-2\\\\3(a^2-2)+7a=4\\3a^2-6+7a-4\\3a^2+7a-10=0\\D=49+120=169=13^2\\a_1=\frac{-7-13}{6}=-\frac{10}{3} \\\\a_2=\frac{-7+13}{6}= 1\\\\\left \{ {{x+\frac{1}{x}=-\frac{10}{3} } \atop {x+\frac{1}{x}=1 }} \right. \\\\x+\frac{1}{x}=-\frac{10}{3} \ (*3x)\\3x^2+3=-10x\\3x^2+10x+3=0\\D=100-36=64=8^2\\x_1=\frac{-10-8}{6}= -3\\x_2=\frac{-10+8}{6}=-\frac{1}{3}$

$x+\frac{1}{x}=1 \ (*x)\\x^2+1=x\\x^2-x+1=0\\D=1-4=-3\\x \in \emptyset$

ОТВЕТ: x1=-3

             x2=-1/3