toi ko bt

\(x^2+y^2=0\)

Mà \(x^2\ge0;y^2\ge0\)nên \(x^2+y^2\ge0\)

(Dấu "="\(\Leftrightarrow x=y=0\))

Ta có \(\frac{1}{P}=\frac{\left(x+yz\right)\left(y+zx\right)\left(z+xy\right)^2}{x^3y^3}=\frac{x+yz}{y}\cdot\frac{y+zx}{x}\cdot\frac{\left(z+xy\right)^2}{x^2y^2}\)

\(=\left(\frac{x}{y}+z\right)\left(\frac{y}{x}+z\right)\left(\frac{z}{xy}+1\right)^2=\left[1+\left(\frac{x}{y}+\frac{x}{y}\right)z+x^2\right]\left(\frac{z}{xy}+1\right)^2\ge\left(1+2x+x^2\right)\)\(\left[\frac{4x}{\left(x+y\right)^2}+1\right]^2\)\(=\left(z+1\right)^2\left[\frac{4z}{\left(z-1\right)^2}+1\right]^2=\left[\frac{4z\left(z+1\right)}{\left(z-1\right)^2}+1\right]^2=\left[6+\frac{12}{z-1}+\frac{8}{\left(z-1\right)^2}+z-1\right]^2\)

\(=\left[6+\frac{12}{z-1}+\frac{3\left(z-1\right)}{4}+\frac{8}{\left(z-1\right)^2}+\frac{z-1}{8}+\frac{z-1}{8}\right]\)

Áp dụng BĐT Cosi ta có:

\(\frac{1}{P}\ge\left[6+2\sqrt{\frac{12}{z-1}\cdot\frac{3\left(z-1\right)}{3}}+3\sqrt[3]{\frac{8}{\left(z-1\right)^2}\cdot\frac{z-1}{8}\cdot\frac{z-1}{8}}\right]^2=\frac{729}{4}\)

\(\Rightarrow P\le\frac{4}{729}\). dấu "=" xảy ra <=> \(\hept{\begin{cases}x=y=2\\z=5\end{cases}}\)

a.

\(\left\{{}\begin{matrix}\left(x-1\right)^2-\left(y+1\right)^2=0\\x+3y-5=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1-y-1\right)\left(x-1+y+1\right)=0\\x+3y-5=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y-2\right)\left(x+y\right)=0\\x+3y-5=0\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}x-y-2=0\\x+3y-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{11}{4}\\y=\dfrac{3}{4}\end{matrix}\right.\)

TH2: \(\left\{{}\begin{matrix}x+y=0\\x+3y-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{2}\\y=\dfrac{5}{2}\end{matrix}\right.\)

b.

\(\left\{{}\begin{matrix}xy-2x-y+2=0\\3x+y=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\left(y-2\right)-\left(y-2\right)=0\\3x+y=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(y-2\right)=0\\3x+y=8\end{matrix}\right.\)

TH1:

\(\left\{{}\begin{matrix}x-1=0\\3x+y=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=5\end{matrix}\right.\)

TH2:

\(\left\{{}\begin{matrix}y-2=0\\3x+y=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)

\(ĐK:x\ne y;x\ne-y;x^2+xy+y^2\ne0;x^2-xy+y^2\ne0\)

\(A=\dfrac{x^2-xy+y^2}{x^2+xy+y^2}\cdot\left[1:\dfrac{\left(x^3+y^3\right)\left(x^2+y^2\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2+y^2\right)}\right]\\ A=\dfrac{x^2-xy+y^2}{x^2+xy+y^2}\cdot\dfrac{\left(x-y\right)\left(x+y\right)\left(x^2+xy+y^2\right)\left(x^2+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)\left(x^2+y^2\right)}\\ A=x-y=B\)

\(x=0;y=0\Leftrightarrow B=0\)

Giá trị của A không xác định vì \(x=y\) trái với ĐK:\(x\ne y\)

Vậy \(A\ne B\)

\(y^2+2\left(x^2+1\right)=2\left(x+1\right)\)

\(\Leftrightarrow y^2+2\left(x^2+x+1\right)=2\left(x+1\right)\)

\(\Leftrightarrow y^2+2x^2+2x+2=2x+2\)

\(\Leftrightarrow y^2+2x^2=0\)

Vì \(x^2\ge0;y^2\ge0\)

\(\Rightarrow y^2+2x^2\ge0\)

Mà \(y^2+2x^2=0\)

Nên \(\hept{\begin{cases}y^2=0\\2x^2=0\end{cases}}\)

Hay x = y = 0

Sau vài phút cố gắng thì khẳng định đề bài của em bị sai