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0乗は1、xの方は原形です。
問題を間違えましたね。えっと、ab-2|+(1+b)²=0 |ab-2|+(1+b)²=0 ∵|ab-2|≧0(1+b)²≧0 ∴|ab-2|=0(1+b)²=0 ∴正解:b=-1 a=-2 ∴-a+3 b=-1
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