Question

Giải hệ:

\(\left\{\begin{matrix}\frac{\left(a+5\right)\left(b+2\right)}{2}=\frac{ab}{2}+45\\\frac{\left(a-1\right)\left(b-1\right)}{2}=\frac{ab}{2}-15\end{matrix}\right.\)

Answer 1

\(\left\{\begin{matrix}\frac{\left(a+5\right)\left(b+2\right)}{2}=\frac{ab}{2}+45\\\frac{\left(a-1\right)\left(b-1\right)}{2}=\frac{ab}{2}-15\end{matrix}\right.\)\(\Leftrightarrow\left\{\begin{matrix}\frac{ab+2a+5b+10-ab}{2}=45\\\frac{ab-a-b+1-ab}{2}=-15\end{matrix}\right.\)

\(\Leftrightarrow\left\{\begin{matrix}2a+5b+10=90\\-a-b+1=-30\end{matrix}\right.\)\(\Leftrightarrow\left\{\begin{matrix}2a+5b=80\\a+b=31\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}2a+5b=80\\2a+2b=62\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}a=25\\b=6\end{matrix}\right.\)

vậy hệ pt có nghiệm (a,b)=(25;6)