Муодиларо ҳал кунед: \((x^2-x-1)^2-x^3=5\)

\((x^2-x-1)^2-x^3=5\)

\((x^2-x-1)^2-x^3-5=0\)

\((x^2-x-1)^2-x^3-4-1=0\)

\((x^2-x-1)^2-x^3-2^2-1^3=0\)

\(((x^2-x-1)^2-2^2)-(x^3+1^3)=0\)

\(((x^2-x-1)^2-2^2)=(x^2-x-1+2)(x^2-x-1-2)=\)

\(=(x^2-x+1)(x^2-x-3)\)

\((x^3+1^3)=(x+1)(x^2-x\cdot 1+1^2)=\)

\(=(x+1)(x^2-x+1)\)

\((x^2-x+1)(x^2-x-3)-(x+1)(x^2-x+1)=0\)

\((x^2-x+1)(x^2-x-3-x-1)=0\)

\((x^2-x+1)(x^2-2x-4)=0\)

\(x^2-x+1=0\)

\(D=1-4\cdot1\cdot1=-3\)

Ин муодилаи квадратӣ реша надорад.

\(x^2-2x-4=0\)

\(D=4+4\cdot1\cdot4=\)

\(=4+16=20\)

\(x_1=\frac{2+\sqrt{20}}{2}=\)

\(=\frac{2+\sqrt{4\cdot5}}{2}=\)

\(=\frac{2+2\cdot\sqrt{5}}{2}=\)

\(=\frac{2\cdot(1+\sqrt{5})}{2}=\)

\(=1+\sqrt{5}\)

\(x_2=\frac{2-\sqrt{20}}{2}=\)

\(=\frac{2-\sqrt{4\cdot5}}{2}=\)

\(=\frac{2-2\cdot\sqrt{5}}{2}=\)

\(=\frac{2\cdot(1-\sqrt{5})}{2}=\)

\(=1-\sqrt{5}\)

Ҷавоб: \(x_1=1+\sqrt{5}\); \(x_2=1-\sqrt{5}\).