证明:如果b^2=ac,则﹙a+b+c﹚﹙a-b+c﹚﹙a²-b²+c²﹚=a^4 +b^4 +c^4

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证明:如果b^2=ac,则﹙a+b+c﹚﹙a-b+c﹚﹙a²-b²+c²﹚=a^4+b^4+c^4证明:如果b^2=ac,则﹙a+b+c﹚﹙a-b+c﹚﹙a²-

证明:如果b^2=ac,则﹙a+b+c﹚﹙a-b+c﹚﹙a²-b²+c²﹚=a^4 +b^4 +c^4
证明:如果b^2=ac,则﹙a+b+c﹚﹙a-b+c﹚﹙a²-b²+c²﹚=a^4 +b^4 +c^4

证明:如果b^2=ac,则﹙a+b+c﹚﹙a-b+c﹚﹙a²-b²+c²﹚=a^4 +b^4 +c^4
左边=[﹙a+c﹚²-b²﹚]﹙a²-b²+c²﹚
=(a²+2ab+b²-b²)﹙a²-b²+c²﹚
=(a²+2b²+b²-b²)﹙a²-b²+c²﹚
=(a²+b²+b²)﹙a²-b²+c²﹚
=(a²+b²)²-(b²)²
=a^4+2a²b²+b^4-b^4
=a^4+2b^4+b^4-b^4
=a^4+b^4+b^4=右边
即﹙a+b+c﹚﹙a-b+c﹚﹙a²-b²+c²﹚=a^4 +b^4 +c^4

分开呀

左边是式子直接展开就行了 不难

﹙a+b+c﹚﹙a-b+c﹚﹙a²-b²+c²﹚=[(a+c)+b][(a+c)-b]﹙a²-b²+c²﹚
=[(a+c)²-b²](a²-b²+c²)
=(a²+2ac+c²-b²)(a²-b²+c²)
...

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﹙a+b+c﹚﹙a-b+c﹚﹙a²-b²+c²﹚=[(a+c)+b][(a+c)-b]﹙a²-b²+c²﹚
=[(a+c)²-b²](a²-b²+c²)
=(a²+2ac+c²-b²)(a²-b²+c²)
=(a²+2b²+c²-b²)(a²-b²+c²)
=(a²+b²+c²)(a²-b²+c²)=a^4+2a²c²+c^4-b^4=a^4 +b^4 +c^4

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