解:(1)原式=
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=
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;
∵a
2+2a-1=0,
∴a
2+2a=1,
∴原式=
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=1;
(2)由已知可得a+b=-c,c=-a-b,
則2a
2+bc=2a
2+b(-a-b)=(a-b)(a+a+b)=(a-b)(a-c),
同理 2b
2+ac=(b-c)(b-a),2c
2+ab=(c-a)(c-b),
∴原式=
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=
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,
∵b=-a-c,
∴a
2(b-c)-b
2(a-c)+c
2(a-b)
=a
2(-a-2c)-(a+c)
2(a-c)+c
2(a+a+c)
=-a
3-2a
2c-a
3-a
2c+ac
2+c
3+2ac
2+c
3=-2a
3-3a
2c+3ac
2+2c
3=2(c
3-a
3)+3ac(c-a)
=(c-a)(2c
2+5ac+2a
2)
=(c-a)(2c+a)(c+2a)
=(c-a)(2c-b-c)(-a-b+2a)
=(a-b)(b-c)(a-c)
∴原式=
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=
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=1.
分析:(1)先把分式化簡,再代入求值即可;
(2)由已知可得a+b=-c,c=-a-b,則2a
2+bc=2a
2+b(-a-b)=(a-b)(a+a+b)=(a-b)(a-c),同理 2b
2+ac=(b-c)(b-a),2c
2+ac=(c-a)(c-b),代入所求代數(shù)式計算即可.
點評:此題考查分式的化簡求值,難度較大,已知條件的反復(fù)應(yīng)用、因式分解的應(yīng)用都要靈活.