【答案】
分析:利用等價(jià)轉(zhuǎn)化思想將這些方程都轉(zhuǎn)化為與之等價(jià)的代數(shù)方程,通過(guò)求解代數(shù)方程達(dá)到求解該方程的目的.注意對(duì)數(shù)中真數(shù)大于零的特點(diǎn).
(1)要注意對(duì)數(shù)式與指數(shù)式的轉(zhuǎn)化關(guān)系;
(2)利用對(duì)數(shù)運(yùn)算性質(zhì)進(jìn)行轉(zhuǎn)化變形;
(3)注意到兩項(xiàng)的聯(lián)系,利用整體思想先求出整體,進(jìn)一步求出方程的根;
(4)利用對(duì)數(shù)的運(yùn)算性質(zhì)進(jìn)行轉(zhuǎn)化與變形是解決本題的關(guān)鍵.注意對(duì)字母的討論.
解答:解:(1)該方程可變形為2x=(x+a)
2,即x=1-a±
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(當(dāng)a≤
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時(shí)),當(dāng)x=1-a-
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時(shí),x+a=1-
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<0,故舍去.因此該方程的根為x=1-a+
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(當(dāng)a≤
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時(shí)),當(dāng)a>
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時(shí),原方程無(wú)根.
(2)該方程可變形為log
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=log
4
,即
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,整理得x
2-7x=0,解出x=0或者x=7(不滿足真數(shù)大于0,舍去).故該方程的根為x=0.
(3)該方程變形為
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=6,即
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,令
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,則可得出t+
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,解得t=3±2
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=
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,因此x=±2.該方程的根為±2.
(4)原方程等價(jià)于
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,由
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得出ax-1=10x-30,該方程當(dāng)a=10時(shí)沒(méi)有根,當(dāng)a≠10時(shí),x=
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,要使得是原方程的根,需滿足ax-1>0,且x-3>0.解出a∈(
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,10).因此當(dāng)a∈(
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,10)時(shí),原方程的根為x=
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,當(dāng)a∈(-∞,
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]∪[10,+∝)時(shí),原方程無(wú)根.
點(diǎn)評(píng):本題考查代數(shù)方程的求解,注意方程的等價(jià)變形,注意對(duì)數(shù)形式方程的真數(shù)大于零的特征,注意對(duì)所求的根進(jìn)行檢驗(yàn),對(duì)含字母的方程要注意討論.