Câu 3.
Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x+2}=a\\\sqrt{y-3}=b\end{matrix}\right.\) ; \(ĐK:x\ne-2;y\ge3\)
hpt trở thành:
\(\left\{{}\begin{matrix}a+2b=7\\2a-3b=7\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=5\\b=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x+2}=5\\\sqrt{y-3}=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}5x+10=1\\y-3=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{9}{5}\\y=4\end{matrix}\right.\) ( tm )
Vậy nghiệm hpt: \(\left\{{}\begin{matrix}x=-\dfrac{9}{5}\\y=4\end{matrix}\right.\)
5.
Ta có:
\(\dfrac{x}{\sqrt{x^2+3}}=\dfrac{x}{\sqrt{x^2+xy+yz+zx}}=\dfrac{x}{\sqrt{\left(x+y\right)\left(x+z\right)}}=\sqrt{\dfrac{x}{x+y}.\dfrac{x}{x+z}}\le\dfrac{1}{2}\left(\dfrac{x}{x+y}+\dfrac{x}{x+z}\right)\)
Tương tự:
\(\dfrac{y}{\sqrt{y^2+3}}\le\dfrac{1}{2}\left(\dfrac{y}{x+y}+\dfrac{y}{y+z}\right)\) ; \(\dfrac{z}{\sqrt{z^2+3}}\le\dfrac{1}{2}\left(\dfrac{z}{x+z}+\dfrac{z}{y+z}\right)\)
Cộng vế:
\(P\le\dfrac{1}{2}\left(\dfrac{x+y}{x+y}+\dfrac{y+z}{y+z}+\dfrac{z+x}{z+x}\right)=\dfrac{3}{2}\)
\(P_{max}=\dfrac{3}{2}\) khi \(x=y=z=1\)