我的解法, 也適用於你新出的延伸題:令a=100, b=11, 根號裏麵分子是
a^4+b^4+(a+b)^4=(a^2+b^2)^2-2a^2b^2+(a^2+b^2+2ab)^2=(a^2+b^2)^2-2a^2b^2+(a^2+b^2)^2+4a^2b^2+4ab(a^2+b^2)=2(a^2+b^2+ab)^2, 所以除以2後開方就是原式,即(a^2+b^2+ab)把原數代回即得100^2+11^2+100*11=11221
我的解法, 也適用於你新出的延伸題:令a=100, b=11, 根號裏麵分子是
a^4+b^4+(a+b)^4=(a^2+b^2)^2-2a^2b^2+(a^2+b^2+2ab)^2=(a^2+b^2)^2-2a^2b^2+(a^2+b^2)^2+4a^2b^2+4ab(a^2+b^2)=2(a^2+b^2+ab)^2, 所以除以2後開方就是原式,即(a^2+b^2+ab)把原數代回即得100^2+11^2+100*11=11221
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