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第 12 週 · 聯立方程與變分 · 第 3 節

分式方程 · 化為二次，驗根去增根

未知數跑到分母上的方程 —— 解法只有一招：兩邊同乘公分母，化成二次方程來解。但天底下最坑的一步在最後 —— 必須驗根去增根。配滿圖例從零講透。

▶ 第一關： 認識分式方程 + 去分母 + 圈禁區
ordinary equation2x + 1 = 7x sits on top — solve directlyfractional equation6x= 2x hides in the DENOMINATOR
multiply BOTH sides by the common denominator x(x−1)6x+6x−1= 5×x(x−1)→no more fractions6(x−1) + 6x = 5·x(x−1) ← a clean polynomial equation
for 6/x + 6/(x−1) = 5 : forbidden zone x = 0 and x = 1x=0 ✗x=1 ✗−12solutions allowed everywhere else
▶ 第二關： 化為二次方程來解（每項都乘）
① clear denom→② expand→③ ax²+bx+c= 0→④ solvefractional equation → one quadratic to crack
6/x + 6/(x−1) = 5 step by step6(x−1) + 6x = 5x(x−1)(multiply by x(x−1))↓12x − 6 = 5x² − 5x↓5x² − 17x + 6 = 0↓(5x − 2)(x − 3) = 0 → x = 2/5 or 3
every term gets × x(x−3) — including the 12/x+2/(x−3)=1most-forgotten term!multiply all three
▶ 第三關： 驗根去增根（本節核心鐵律）
ORIGINAL equationx = 0,1 forbidden×denom(erases the limit)polynomial equationno limit anymore⇒ a root that hits x=0 or 1 can sneak in — an EXTRANEOUS rootit fits the polynomial but breaks the original fraction
verify every root in the ORIGINAL denominatorsolved: x = 3↓ put back into x − 33 − 3 = 0 ✗ denominator diesx = 3 is EXTRANEOUS → reject → no solution
x² = 1 → two candidate roots, check eachx = 1x−1 = 0 ✗ rejectx = −1x−1 = −2 ✓ keepfinal answer : x = −1 only
⚔ BOSS： 分式方程綜合（含必出增根）
master plan:① markforbidden→② cleardenom→③ solvequadratic→④ VERIFY → dropextraneous roots