\(\left(x-y-5\right)^2+\left|2x-3y\right|=0\)
\(\left\{{}\begin{matrix}\left(x-y-5\right)^2\ge0\forall x;y\\\left|2x-3y\right|\ge0\forall x;y\end{matrix}\right.\)
\(\Rightarrow\left(x-y-5\right)^2+\left|2x-3y\right|\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left(x-y-5\right)^2=0\\\left|2x-3y\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x-y-5=0\\2x-3y=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x-y=5\\2x-3y=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x-2y=10\Rightarrow2x=10+2y\\2x-3y=0\Rightarrow2x=3y\end{matrix}\right.\)
\(\Rightarrow10+2y=3y\Rightarrow y=10\)
\(\Rightarrow x-10=5\Rightarrow x=10+5=15\)
Mình k chép lại đề nha!
=> x-y-5=0 => x-y=5
Và 2x-3y=0 => 2x=3y => \(\dfrac{x}{3}=\dfrac{y}{2}\)
Ap dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{3}=\dfrac{y}{2}=\dfrac{x-y}{3-2}=\dfrac{5}{1}=5\)
Suy ra x/3=5=>x=15
y/2=5=>y=10
Vậy x=15 ; y=10
