Question

cho a^2-2b+1 = 0 ; b^2-2c+1 ; c^2-2a + 1 = 0.
Tính P = ( a-2 )^201 + (b-2)^202 + (c-2)^203

Answer 1

Từ \(a^2-2b+1=0;b^2-2c+1=0;c^2-2a+1=0\)

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

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

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

\(\Rightarrow\hept{\begin{cases}a-1=0\\b-1=0\\c-1=0\end{cases}\Rightarrow a=b=c=1}\)

\(\Rightarrow P=\left(1-2\right)^{201}+\left(1-2\right)^{202}+\left(1-2\right)^{203}=-1+1-1=-1\)