分析:(1)首先根據(jù)去括號(hào)法則,去掉原式的括號(hào),再合并同類項(xiàng),然后,把A,B所表示的代數(shù)式代入,最后去括號(hào)、合并同類項(xiàng)即可,(2)首先根據(jù)去括號(hào)法則,去掉原式的括號(hào),再合并同類項(xiàng),然后,把A,B所表示的代數(shù)式代入,經(jīng)過去括號(hào),合并同類項(xiàng)進(jìn)行化簡(jiǎn),最后把a(bǔ)的值代入化簡(jiǎn)后的代數(shù)式即可求出結(jié)果.
解答:解:(1)∵A=2a
2-a,B=a
2-2a+1,
∴A-2(A-B)-3=-A+2B-3
=-2a
2+a-2(a
2-2a+1)-3
=-3a-1,
(2)∵A=2a
2-a,B=a
2-2a+1,
∴當(dāng)a=
-時(shí),
原式=-A+2B+3
=-3a-1+6
=-3×(
-)+5
=6.
點(diǎn)評(píng):本題主要考查整式的化簡(jiǎn)求值,合并同類項(xiàng),去括號(hào)法則等知識(shí)點(diǎn),關(guān)鍵在于認(rèn)真、正確地進(jìn)行計(jì)算.