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

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

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

Bài a hình như sai đề rồi bạn.

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

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

Sao không ai giúp hết vậy!

Câu 3 là (1+1/x)(1+1/y) nha

Mà ko cần làm câu này đâu giúp mình 2 câu 1 và 2 thôi nhá

\(2x+3y=1\Rightarrow y=\frac{1-2x}{3}\)

Do \(x;y\ge0\Rightarrow0\le x\le\frac{1}{2}\)

\(A=x^2+3\left(\frac{1-2x}{3}\right)^2=x^2+\frac{1}{3}\left(4x^2-4x+1\right)=\frac{7}{3}x^2-\frac{4}{3}x+\frac{1}{3}\)

\(A=\frac{7}{3}\left(x-\frac{2}{7}\right)^2+\frac{1}{7}\ge\frac{1}{7}\)

\(\Rightarrow A_{min}=\frac{1}{7}\) khi \(x=\frac{2}{7};y=\frac{1}{7}\)

Mặt khác \(A=\frac{1}{3}x\left(7x-4\right)+\frac{1}{3}\)

Do \(x\le\frac{1}{2}\Rightarrow7x-4< 0\Rightarrow x\left(7x-4\right)\le0\)

\(\Rightarrow A\le\frac{1}{3}\Rightarrow A_{max}=\frac{1}{3}\) khi \(x=0;y=\frac{1}{3}\)

Câu 2:

\(A-4=2x+3y\Rightarrow\left(A-4\right)^2=\left(2x+3y\right)^2\)

\(\left(A-4\right)^2\le\left(2^2+3^2\right)\left(x^2+y^2\right)=676\)

\(\Rightarrow-26\le A-4\le26\)

\(\Rightarrow-22\le A\le30\)

\(A_{max}=30\) khi \(\left\{{}\begin{matrix}x=4\\y=6\end{matrix}\right.\)

\(A_{min}=-22\) khi \(\left\{{}\begin{matrix}x=-4\\y=-6\end{matrix}\right.\)

\(A^2=\left(2x+3y\right)^2\le\left(2^2+3^2\right)\left(x^2+y^2\right)=13^2\)

\(\Rightarrow A\le13\Rightarrow A_{max}=13\) khi \(\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)