Gọi chiều rộng , chiều dài ban đầu của HCN là a ; b \(\left(a;b>0\right)\)
Ta có PT :
\(\left\{{}\begin{matrix}2\left(a+b\right)=200\\2\left(a+b\right)+4=2\left[a+\dfrac{1}{10}b+b-\dfrac{1}{10}a\right]\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=100\\200+4=2\left[100+\dfrac{1}{10}\left(b-a\right)\right]\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=100\\100+\dfrac{1}{10}\left(b-a\right)=102\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=100\\b-a=20\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=40\\b=60\end{matrix}\right.\)
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