Question

tìm x

|x+23|^2002+|y-12|^123=0

Answer 1

|x+23|^2002|+|y-12|^123=0

<=>|x+23|^2002\(\ge\) 0

      \(\left|y-12\right|^{^{123}}\ge0\)

=>\(\left|x+23\right|^{2002}+\left|y-12\right|^{123}\ge0\)

=>x+23=0=>x=-23

=>y-12=0=>y=12

Answer 2

\(\left|x+23\right|^{2002}+\left|y-12\right|^{123}=0\)

\(\Rightarrow\hept{\begin{cases}\left|x+23\right|^{2002}=0\\\left|y-12\right|^{123}=0\end{cases}}\Rightarrow\hept{\begin{cases}x+23=0\\y-12=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-23\\y=12\end{cases}}\)

Vậy \(x=-23;y=12\)

Answer 3

Vì \(\left|x+23\right|^{2002}\ge0;\left|y-12\right|^{123}\ge0\forall x;y\)

\(\Rightarrow\left|x+23\right|^{2002}+\left|y-12\right|^{123}\ge0\forall x;y\)

Mà đề lại cho \(\left|x+23\right|^{2002}+\left|y-12\right|^{123}=0\)

\(\Leftrightarrow\left|x+23\right|^{2002}=0;\left|y-12\right|^{123}=0\)

\(\Rightarrow x=-23;y=12\)

Answer 4

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