Question

помогите мне пожалуйста

Answer 1

$x=psi\=psi^2+frac{psi^2}{psi^2-2psi+1}=8 \\\=dfrac{psi^2}{psi^2-2psi+1}+dfrac{psi^2}{1}=8 \\\dfrac{psi^2}{psi^2-2psi+1}+dfrac{psi^2(psi^2-2psi+1)}{psi^2-2psi+1}=8 \\\fdrac{psi^2+psi^2(psi^2-2psi+1)}{psi^2-2psi+1}=8 \\\dfrac{psi^4-2psi^3+2psi^2}{psi^2-2psi+1}=8 \\\dfrac{psi^4-2psi^3+2psi^2}{psi^2-2psi+1}(psi^2-2psi+1)=8(psi^2-2psi+1) \psi^4-2psi^3+2psi^2=8(psi^2-2psi+1) \psi^4-2psi^3+2psi^2=8psi^2-16psi+8 \psi^4-2psi^3+2psi^2-(8psi^2-16psi+8)=0 \psi^4-2psi^3-6psi^2+16psi-8=0 \(psi-2)^2(psi^2+2psi-2)=0 \\\boxed{ dfrac{psi=2}{dfrac{psi=sqrt{3}-1}{psi=-1-sqrt{3}} } } \boxed{textbf{Good day}}$

Answer 2

$x^2+( frac{x}{x-1} )^2=8$
$x^2+frac{x^2}{x^2-2x+1} =8$
$x^2(x^2-2x+1)+x^2=8(x^2-2x+1)$
$x^4-2x^3-6x^2+16x-8=0$
$(x-2)^2(x^2+2x-2)=0$
$x_{1}=2$
$x^2+2x-2=0$
$D = 12$
$x_{2,3}= -1б sqrt{3}$