Question

8．已知x=2-$\sqrt{2}$，則代數(shù)式x²-2x-1的值為1-2$\sqrt{2}$．

Answer 1

分析 先對(duì)原代數(shù)式進(jìn)行恰當(dāng)?shù)幕��?jiǎn)，然后代入求值即可．

解答 解：原式=x²-2x+1-2
=（x-1）²-2
=（2-$\sqrt{2}$-1）²-2
=（1-$\sqrt{2}$）²-2
=1+2-2$\sqrt{2}$-2
=1-2$\sqrt{2}$．

故答案為：1-2$\sqrt{2}$．

點(diǎn)評(píng) 本題考查了二次根式的化簡(jiǎn)求值，解答本題的關(guān)鍵在于對(duì)原式進(jìn)行恰當(dāng)?shù)幕��?jiǎn)并代入求值．