Mathematics
Senior High
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(2)の解き方を教えてください
2. 次の分数式の値を求めよ。
bc+ca
ab
=
betca
ab
ca+ab
bc
(1) ab+bc+ca≠0 のとき
catab
bc
[bc+ca = abk
catab = bc k
ab+bc = ca k
-
ab + bc
ca
_ ab + bc = k ck +0) kzić
ca
辺々を加えると
2(ab+bc+ca)=f(ab+be+ca)
(2) ab+bc+ca=0のとき
[解] ab+bc+ca = 0 より
bc+ca=-ab
ca+ab = -bc
ab+bc=-ca
ab + bc+ca 074
k=2
(与式)=2
辺々を加えると
2(ab+be+ca) == (ab+be+ca)
ab+be+ca=07₁4
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