Question

C= 5x¹⁰ - y¹⁵ +2007 biết (x+1)²⁰⁰⁶ +(y-1)²⁰⁰⁸ =0

Answer 1

\(\left(x+1\right)^{2006}+\left(y-1\right)^{2008}=0\)

\(\left\{{}\begin{matrix}\left(x+1\right)^{2006}\ge0\\\left(y-1\right)^{2008}\ge0\end{matrix}\right.\)

\(\Rightarrow\left(x+1\right)^{2006}+\left(y-1\right)^{2008}\ge0\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}\left(x+1\right)^{2006}=0\\\left(y-1\right)^{2008}=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)

Thay vào C ta có:

\(C=5.\left(-1\right)^{10}-1^{15}+2007\)

\(=5-1+2007=2011\)

Answer 2

(x+1)²⁰⁰⁶+(y-1)²⁰⁰⁸=0

=> (x+1)²⁰⁰⁶=(y-1)²⁰⁰⁸=0

=>x+1=y-1=0

=>x=-1 và y=1

C=5x¹⁰-y¹⁵+2007=5.(-1)¹⁰-1¹⁵+2007=2011