Question

Cho \(x=\frac{1}{2}\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}\). Tính:

\(M=\left(4x^5+4x^4-x^3+1\right)^{19}+\left(\sqrt{4x^5+4x^4-5x^3+5x+3}\right)^3+\left(\frac{1-\sqrt{2}}{\sqrt{2x^2+2x}}\right)^{2016}\)

Answer 1

\(1-\sqrt{2}x\) nha

Answer 2

\(x=\frac{1}{2}\left(\sqrt{2}-1\right)\)

\(\Leftrightarrow2x=\sqrt{2}-1\Leftrightarrow4x^2=3-2\sqrt{2}=1-4.\frac{1}{2}\left(\sqrt{2}-1\right)=1-4x\)

\(\Leftrightarrow4x^2+4x-1=0\)

\(\left[x^3\left(4x^2+4x-1\right)+1\right]^{19}=1^{19}=1\)

\(\sqrt{x^3\left(4x^2+4x-1\right)-x\left(4x^2+4x-1\right)+4x^2+4x-1+4}^3=\sqrt{4}^3=8\)

\(\frac{1-\sqrt{2}x}{\sqrt{\frac{1}{2}\left(4x^2+4x-1\right)+\frac{1}{2}}}=\frac{1-\sqrt{2}x}{\sqrt{\frac{1}{2}}}=\sqrt{2}-2x=\sqrt{2}-\left(\sqrt{2}-1\right)=1\)

\(M=1+8+1=10\)