Question

Cho \(x^2+2x-2=0\). Tính giá trị của biểu thức \(M=x^4+16x+2007\)

Answer 1

x^2+2x-2=0

=>\(\left[{}\begin{matrix}x=\dfrac{-1+\sqrt{5}}{2}\\x=\dfrac{-1-\sqrt{5}}{2}\end{matrix}\right.\)

Nếu \(x=\dfrac{-1+\sqrt{5}}{2}\) thì \(M=\left(\dfrac{-1+\sqrt{5}}{2}\right)^4+16\cdot\dfrac{-1+\sqrt{5}}{2}+2007\)

\(=\left(\dfrac{6-2\sqrt{5}}{4}\right)^2-8+8\sqrt{5}+2007\)

\(=\left(\dfrac{3-\sqrt{5}}{2}\right)^2+1999+8\sqrt{5}\)

\(=\dfrac{14-6\sqrt{5}}{4}+1999+8\sqrt{5}\)

\(=3.5-1.5\sqrt{5}+8\sqrt{5}+1999=2002.5+6.5\sqrt{5}\)

Nếu \(x=\dfrac{-1-\sqrt{5}}{2}\) thì \(M=\left(\dfrac{6+2\sqrt{5}}{4}\right)^2+16\cdot\dfrac{-\sqrt{5}-1}{2}+2007\)

\(=\dfrac{14+6\sqrt{5}}{4}-8\sqrt{5}+8+2007\)

\(=\dfrac{4037}{2}+\dfrac{-13}{2}\sqrt{5}\)