Question

Cho a,b>0 thoả mãn a³+b³=3ab-1

CMR: a²⁰¹⁸+b²⁰¹⁹=2

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Answer 1

     \(a^3+b^3=3ab-1\)

\(\Rightarrow a^3+b^3+1-3ab=0\)

\(\Rightarrow\left(a+b\right)^3+1-3ab\left(a+b\right)-3ab=0\)

\(\Rightarrow\left(a+b+1\right)\left(a^2+2ab+b^2-a-b+1\right)-3ab\left(a+b\right)=0\)

\(\Rightarrow\left(a+b+1\right)\left(a^2-ab+b^2-a-b+1\right)=0\)

Mà \(a,b>0\Rightarrow a+b+1>0\)

\(\Rightarrow a^2-ab+b^2-a-b+1=0\)

\(\Rightarrow2a^2-2ab+2b^2-2a-2b+2=0\)

\(\Rightarrow\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2=0\)

\(\Rightarrow a=b=1\Rightarrow a^{2018}+b^{2019}=1+1=2\)