Đề sai à? check lại đề đi

\(a,\text{Đặt }\left\{{}\begin{matrix}S=y+z\\P=yz\end{matrix}\right.\\ HPT\Leftrightarrow\left\{{}\begin{matrix}\left(y+z\right)^2-2yz+x^2=8\\x\left(y+z\right)+yz=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}S^2-2P+x^2=8\\Sx+P=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}S^2-2\left(4-Sx\right)+x^2=8\\P=4-Sx\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}S^2+2Sx+x^2-16=0\left(1\right)\\P=4-Sx\end{matrix}\right.\\ \left(1\right)\Leftrightarrow\left(S+x-4\right)\left(S+x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}S=-x+4\Rightarrow P=\left(x-2\right)^2\\S=-x-4\Rightarrow P=\left(x+2\right)^2\end{matrix}\right.\)

Mà y,z là nghiệm của hệ nên \(S^2-4P\ge0\Leftrightarrow\left[{}\begin{matrix}\left(4-x\right)^2\ge4\left(x-2\right)^2\\\left(-4-x\right)^2\ge4\left(x+2\right)^2\end{matrix}\right.\Leftrightarrow-\dfrac{8}{3}\le x\le\dfrac{8}{3}\)

\(c,P=\dfrac{x^2-x^2+8xy-16y^2}{x^2+4y^2}=\dfrac{8\left(\dfrac{x}{y}\right)-16}{\left(\dfrac{x}{y}\right)^2+4}\)

Đặt \(\dfrac{x}{y}=t\)

\(\Leftrightarrow P=\dfrac{8t-16}{t^2+4}\Leftrightarrow Pt^2+4P=8t-16\\ \Leftrightarrow Pt^2-8t+4P+16=0\)

Với \(P=0\Leftrightarrow t=2\)

Với \(P\ne0\Leftrightarrow\Delta'=16-P\left(4P+16\right)\ge0\)

\(\Leftrightarrow-P^2-4P+4\ge0\Leftrightarrow-2-2\sqrt{2}\le P\le-2+2\sqrt{2}\)

Vậy \(P_{max}=-2+2\sqrt{2}\Leftrightarrow t=\dfrac{4}{P}=\dfrac{4}{-2+2\sqrt{2}}=2+\sqrt{2}\)

\(\Leftrightarrow\dfrac{x}{y}=2+2\sqrt{2}\)

\(\Leftrightarrow-4x-3+5x+2=0\\ \Leftrightarrow x-1=0\Leftrightarrow x=1\)

Áp dụng Viét: \(\left\{{}\begin{matrix}x_1+x_2=3\\x_1x_2=-2\end{matrix}\right.\)

\(A=x_1+x_2=3\\ B=x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2=3^2-2\left(-2\right)=9+4=13\\ C=x_1^3+x_2^3=\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)=3^3-3\left(-2\right)\cdot3=27-18=9\\ D=x_1^4+x_2^4=\left(x_1^2+x_2^2\right)^2-2x_1^2x_2^2=13^2-2\left(-2\right)^2=161\)