a.
ĐKXĐ: \(x>0\)
Xét pt đầu:
\(\dfrac{1}{\sqrt[]{x}}+\dfrac{y}{x}=\dfrac{2\sqrt[]{x}}{y}+2\)
\(\Leftrightarrow\dfrac{y+\sqrt[]{x}}{x}=\dfrac{2\left(y+\sqrt[]{x}\right)}{y}\)
\(\Rightarrow\left[{}\begin{matrix}y+\sqrt[]{x}=0\\y=2x\end{matrix}\right.\)
- Với \(y+\sqrt[]{x}=0\Rightarrow y=-\sqrt[]{x}< 0\)
\(\Rightarrow\left\{{}\begin{matrix}y\left(\sqrt[]{x^2+1}-1\right)< 0\\\sqrt[]{3x^2+3}>0\end{matrix}\right.\) hệ vô nghiệm
- Với \(y=2x\)
\(\Rightarrow2x\left(\sqrt[]{x^2+1}-1\right)=\sqrt[]{3\left(x^2+1\right)}\)
\(\Leftrightarrow2x\sqrt[]{x^2+1}-2x=\sqrt[]{3\left(x^2+1\right)}\)
\(\Leftrightarrow x\left(\sqrt[]{x^2+1}-2\right)+\sqrt[]{x^2+1}\left(x-\sqrt[]{3}\right)=0\)
\(\Leftrightarrow\dfrac{x\left(x+\sqrt[]{3}\right)\left(x-\sqrt[]{3}\right)}{\sqrt[]{x^2+1}+2}+\sqrt[]{x^2+1}\left(x-\sqrt[]{3}\right)=0\)
\(\Leftrightarrow\left(x-\sqrt[]{3}\right)\left(\dfrac{x\left(x+\sqrt[]{3}\right)}{\sqrt[]{x^2+1}+2}+\sqrt[]{x^2+1}\right)=0\)
\(\Leftrightarrow x=\sqrt[]{3}\Rightarrow y=2\sqrt[]{3}\)
b.
Không mất tính tổng quát, giả sử \(z=min\left\{x;y;z\right\}\Rightarrow x\ge z\Rightarrow4x\ge4z\)
\(\Rightarrow z^3-3z^2+5z+1\ge y^3-3y^2+5y+1\)
\(\Rightarrow z^3-y^3-3z^2+3y^2+5z-5y\ge0\)
\(\Leftrightarrow\left(z-y\right)\left(z^2+yz+y^2\right)-\left(z-y\right)\left(3z+3y\right)+5\left(z-y\right)\ge0\)
\(\Leftrightarrow\left(z-y\right)\left(z^2+yz+y^2-3z-3y+5\right)\ge0\)
\(\Leftrightarrow\left(z-y\right)\left[\dfrac{1}{2}\left(y+z-2\right)^2+\dfrac{1}{2}\left(y-1\right)^2+\dfrac{1}{2}\left(z-1\right)^2+2\right]\ge0\)
\(\Leftrightarrow z-y\ge0\Rightarrow z\ge y\) (1)
Mặt khác cũng do giả sử \(z=min\left\{x;y;z\right\}\Rightarrow z\le y\) (2)
(1);(2) \(\Rightarrow z=y\)
Lý luận tương tự ta được \(z=x\)
\(\Rightarrow x=y=z\)
Thay vào pt đầu:
\(\Rightarrow x^3-3x^2+5x+1=4x\)
\(\Leftrightarrow x^3-3x^2+x+1=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-2x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=y=z=1\\x=y=z=1-\sqrt{2}\\x=y=z=1+\sqrt{2}\end{matrix}\right.\)