Question

Cho \(x=\sqrt{2}-1\). Tính P = \(x^8+x^6-x^2+3\)

Answer 1

Có \(\left\{{}\begin{matrix}x^2=\left(\sqrt{2}-1\right)^2=3-2\sqrt{2}\\x^3=\left(\sqrt{2}-1\right)^3=2\sqrt{2}-6+3\sqrt{2}-1=5\sqrt{2}-7\end{matrix}\right.\)

Có \(P=x^8+x^6-x^2+3\)

P=\(x^2\left(x^6+x^4-1\right)+3\)

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

=\(x^2\left[\left(5\sqrt{2}-7+1\right)\left(5\sqrt{2}-7-1\right)+\left(3-2\sqrt{2}\right)^2\right]+3\)

=\(x^2\left[\left(5\sqrt{2}-6\right)\left(5\sqrt{2}-8\right)+17-12\sqrt{2}\right]+3\)

=\(x^2\left(50-40\sqrt{2}-30\sqrt{2}+48+17-12\sqrt{2}\right)+3\)

=\(\left(3-2\sqrt{2}\right)\left(115-82\sqrt{2}\right)+3\)

=\(345-246\sqrt{2}-230\sqrt{2}+328+3\)

=\(676-476\sqrt{2}\)

Answer 2

Lời giải:

\(x=\sqrt{2}-1\Rightarrow x+1=\sqrt{2}\)

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

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

Do đó:

\(P=x^8+x^6-x^2+3=(x^8-1)+(x^6-x^2)+4\)

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

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

\(=(4x^2-4x)(5x^2-4x+2)+4\) (theo $(2)$)

\(=(4-8x-4x)(5-10x-4x+2)+4\) (theo $(1)$)

\(=(4-12x)(7-14x)+4=28(1-3x)(1-2x)+4\)

\(=28(1-5x+6x^2)+4=28(1-5x+6-12x)+4\)

\(=28(7-17x)+4=200-476x=200-476(\sqrt{2}-1)=676=476\sqrt{2}\)

Answer 3

@Nguyễn Việt Lâm

Answer 4

@Vũ Minh Tuấn

Answer 5

@HISINOMA KINIMADO

Answer 6

@Lê Thị Thục Hiền

Answer 7

@Nguyễn Huy Tú

Answer 8

@Akai Haruma

Answer 9

@Ace legona

Answer 10

@Mysterious Person

Answer 11

@Võ Đông Anh Tuấn

Answer 12

@soyeon_Tiểubàng giải