Question

Иррациональное уравнение решите, пожалуйста!!! Используя замену t
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Answer 1

$frac{1}{1- sqrt{1-x} } -frac{1}{1+sqrt{1-x}} = frac{2+2sqrt2}{sqrt{1-x}} ; , ; ; ; ; ; left { {{1-sqrt{1-x}ne 0} atop {1-x textgreater 0}} right. \\t=sqrt{1-x}; ; ,; ; frac{1}{1-t}+frac{1}{1+t}=frac{2+2sqrt2}{t}\\frac{1+t+1-t}{1-t^2}-frac{2+2sqrt2}{t}=0; ;; ; frac{2}{1-t^2}-frac{2+2sqrt2}{t}=0\\frac{2t-2-2sqrt2+2t^2-2sqrt2t^2}{t(1-t^2)}=0\\frac{(2-2sqrt2)t^2+2t-(2+2sqrt2)}{t(1-t^2)}=0\\frac{(1-sqrt2)t^2+t-(1+sqrt2)}{t(1-t^2)}=0; ; to$

$left { {{(1-sqrt2)t^2+t-(1+sqrt2)=0} atop {t(1-t^2)ne 0}} right. \\D=1-4(1-sqrt2)(1+sqrt2)=1-4(1-2)=5\\t_1=frac{-1-sqrt5}{2(1-sqrt2)}=frac{(-1-sqrt5)(1+sqrt2)}{2(1-2)}=frac{-1-sqrt2-sqrt5-sqrt{10}}{-2}=frac{1+sqrt2+sqrt5+sqrt{10}}{2}\\t_2=frac{-1+sqrt5}{2(1-sqrt2)}=frac{(-1+sqrt5)(1+sqrt2)}{2(1-2)}=frac{-1-sqrt2+sqrt5+sqrt{10}}{-2}=frac{1+sqrt2-sqrt5-sqrt{10}}{2}$

 $sqrt{1-x}=frac{1+sqrt2+sqrt5+sqrt{10}}{2}; ,$ $x=1-frac{(1+sqrt2+sqrt5+sqrt{10})^2}{4}$ 

$sqrt{1-x}=frac{1+sqrt2-sqrt5-sqrt{10}}{2}; ; to ; ; x=1-frac{(1+sqrt2-sqrt5-sqrt{10})^2}{4}$

Answer 2

спасибо!!!