Question

Biết a,b>0; a+b=a^2+b^2=a^3+b^3. Tính a^2015+b^2015

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Answer 1

Có: \(a+b=a^2+b^2=a^3+b^3\)

\(\Rightarrow a+b+a^3+b^2=2\left(a^2+b^2\right)\)

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

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

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

Vì: \(a>0;\left(1-a\right)^2\ge0\)

=> \(a\left(1-a\right)^2\ge0\)

Vì: \(b>0;\left(1-b\right)^2\ge0\)

=> \(b\left(1-b\right)^2\ge0\)

Do đó:

\(\left(1\right)\Leftrightarrow\begin{cases}a\left(1-a\right)^2=0\\b\left(1-b\right)^2=0\end{cases}\)\(\Leftrightarrow\begin{cases}1-a=0\\1-b=0\end{cases}\)\(\Leftrightarrow a=b=1\)

Khi đó; \(a^{2015}+b^{2015}=1^{2015}+1^{2015}=2\)