Question

\(x+y+z>11\)

VA \(8x+9y+10z=100\)

TIM X,Y,Z ???

Answer 1

\(\hept{\begin{cases}x+y+z>1\\8x+9y+10z=100\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x+y+z\ge12\\8x+9y+10z=100\end{cases}}\)

\(\Rightarrow y+2z=100-8\left(x+y+z\right)\le100-8\cdot12=4\)

Mặt khác \(y,z\ge1\)suy ra \(3\le y+2z\le4\)\(\Rightarrow y+2z\in\left\{3,4\right\}\)

Nếu \(y+2z=3\Leftrightarrow y=z=1\Rightarrow x\in\left\{\text{Ø}\right\}\)Nếu \(y+2z=4\Leftrightarrow y=2;z=1\Rightarrow x=9\)