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國二關於根號的數學問題

Nov 3rd 2013, 06:09

a√(1-b²)+b√(1-a²)=1

移項→a√(1-b²)=1-b√(1-a²)

平方→a²(1-b²)=1+b²(1-a²)-2b√(1-a²)

移項→2b√(1-a²)=1+b²(1-a²)-a²(1-b²)

整理→2b√(1-a²)=1+b²-a²

平方→4b²(1-a²)=[1+(b²-a²)]²

展開→4b²-4a²b²=1+(a²-b²)²-2(a²-b²)

移項→1+(a²-b²)²-2(a²-b²)+4a²b²-4b²=0

合併→1+(a²+b²)²-2(a²+b²)=0

合併→[1-(a²+b²)]²=0

得解→(a²+b²)=1

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