tex]...

lowmore @lowmore

July 2022 1 6 Report

[tex] x^{2} + \frac{1}{x^{2}} = 2*(x- \frac{1}{x})+2[/tex]

Answers & Comments

AnonimusPro

Verified answer

Делаем замену:
$y= x-\frac{1}{x} 
\\y^2=x^2-2+ \frac{1}{x^2}
\\y^2+2=x^2+ \frac{1}{x^2}$
получаем:
$y^2+2=2y+2 \\y^2-2y=0 \\y(y-2)=0 \\y_1=0 \\y_2=2 \\ \\x- \frac{1}{x}=0
\\x^2-1=0
\\x^2=1
\\x_1=1
\\x_2=-1
\\
\\x- \frac{1}{x} =2
\\x^2-2x-1=0
\\D=4+4=8=(2\sqrt{2})^2
\\x_1= \frac{2+2\sqrt{2}}{2}=\sqrt{2}+1
\\x_2= \frac{2-2\sqrt{2}}{2} =1-\sqrt{2}$
Ответ: $x_1=1;x_2=-1;x_3=\sqrt{2}+1;x_4=1-\sqrt{2}$

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