\(xy\left(x-2\right)\left(y-2\right)=4\)

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

\(\Rightarrow\left(x^2-2x\right)y^2+\left(4x-2x^2\right)y-4=0\)

\(\left(x^2-2x\right)y^2+\left(-2x^2+4xy\right)y-4=0\)

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

\(\Rightarrow y\left(y-2\right)=\frac{4}{x-\left(x-2\right)}\)

\(\left(x-2\right)\ne0\)

\(\Leftrightarrow\orbr{y=\frac{x^2-\sqrt{x^4-4x^3+8x^2-8x-2x}}{x^2-2x}}\)

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1/

Ta có: \(x-y=xy\Rightarrow x=xy+y=y\left(x+1\right)\Rightarrow x:y=x+1\left(y\ne0\right)\)

Mà x - y =  x:y

\(\Rightarrow x-y=x+1\Rightarrow-y=1\Rightarrow y=-1\)

Thay y = -1 vào x - y = xy ta được:

\(x-\left(-1\right)=x.\left(-1\right)\Rightarrow x+1=-x\Rightarrow2x=-1\Rightarrow x=\frac{-1}{2}\)

Vậy...

2/ tương tự bài 1    x = 1/2, y = -1

1. (x;y;z) = (2;2;2) . Đó là hpt đối xứng

2.(x;y;z) = (1;1;1) . Đây cũng là hpt đối xứng

A=x^4+y^4-xy\(-\left(x^2y^2+7xy-9\right)\)

A=\(\left(x^2+y^2\right)^2-2x^2y^2-xy\)

A=\(\left(3-xy\right)^2-2x^2y^2-xy\)

A=\(-\left(x^2y^2+7xy-9\right)\)

A=\(-\left(x^2y^2+6xy+9+xy-18\right)\)

A=\(-\left(xy+3\right)^2-xy+18\)

Đến đây đánh giá xy

Có x^2+y^2+xy=3

hay (x+y)^2=3+xy

suy ra xy+3>=0

hay xy>=-3

Như vậy A<=21

Dấu bằng xảy ra khi x=\(\sqrt{3}\),y=\(-\sqrt{3}\)

Chúc bạn học tốt

1.

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y+x^3y+xy^2+xy=-\dfrac{5}{4}\\x^4+y^2+xy\left(1+2x\right)=-\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2+y\right)+xy+xy\left(x^2+y\right)=-\dfrac{5}{4}\\\left(x^2+y\right)^2+xy=-\dfrac{5}{4}\end{matrix}\right.\left(1\right)\)

Đặt \(\left\{{}\begin{matrix}x^2+y=a\\xy=b\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}a+b+ab=-\dfrac{5}{4}\\a^2+b=-\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a-a^2-\dfrac{5}{4}-a\left(a^2+\dfrac{5}{4}\right)=-\dfrac{5}{4}\\b=-a^2-\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a^2-a^3-\dfrac{1}{4}a=0\\b=-a^2-\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-a\left(a^2-a+\dfrac{1}{4}\right)=0\\b=-a^2-\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a\left(a-\dfrac{1}{2}\right)^2=0\\b=-a^2-\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=0\\b=-\dfrac{5}{4}\end{matrix}\right.\\\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=-\dfrac{3}{2}\end{matrix}\right.\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}a=0\\b=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2+y=0\\xy=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{\sqrt[3]{10}}{2}\\y=-\dfrac{5}{2\sqrt[3]{10}}\end{matrix}\right.\)

TH2: \(\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2+y=\dfrac{1}{2}\\xy=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-\dfrac{3}{2}\end{matrix}\right.\)

Kết luận: Phương trình đã cho có nghiệm \(\left(x;y\right)\in\left\{\left(\dfrac{\sqrt[3]{10}}{2};-\dfrac{5}{2\sqrt[3]{10}}\right);\left(1;-\dfrac{3}{2}\right)\right\}\)

2.

\(\left\{{}\begin{matrix}\left(x+1\right)^3-16\left(x+1\right)=\left(\dfrac{2}{y}\right)^3-4\left(\dfrac{2}{y}\right)\\1+\left(\dfrac{2}{y}\right)^2=5\left(x+1\right)^2+5\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+1=u\\\dfrac{2}{y}=v\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u^3-16u=v^3-4v\\v^2=5u^2+4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}u^3-v^3=16u-4v\\4=v^2-5u^2\end{matrix}\right.\)

\(\Rightarrow4\left(u^3-v^3\right)=\left(16u-4v\right)\left(v^2-5u^2\right)\)

\(\Leftrightarrow21u^3-5u^2v-4uv^2=0\)

\(\Leftrightarrow u\left(7u-4v\right)\left(3u+v\right)=0\Rightarrow\left[{}\begin{matrix}u=0\Rightarrow v^2=4\\u=\dfrac{4v}{7}\Rightarrow4=v^2-5\left(\dfrac{4v}{7}\right)^2\\v=-3u\Rightarrow4=\left(-3u\right)^2-5u^2\end{matrix}\right.\) 

\(\Rightarrow...\)