\(pt< =>\left(x-y\right)^2+xy=\left(x-y\right)\left(xy+2\right)+9\)

\(< =>\left(y-x\right)\left(xy+2+y-x\right)+xy+2+y-x-\left(y-x\right)=11\)

\(< =>\left(y-x+1\right)\left(xy+2+y-x\right)-\left(y-x+1\right)=10\)

\(< =>\left(x-y+1\right)\left(x-y-1-xy\right)=10\)

đến đây giải hơi bị khổ =))

ĐKXĐ: x>=0; y>=1; 2y-3x-4>=0

\(\begin{cases} x^3 + x^2 + y^2 - x^2y - xy - y = 0 \quad (1) \\ \sqrt{x} + \sqrt{y - 1} = \sqrt{2y - 3x - 4} \quad (2) \end{cases}\)

(1): \(x^3 + x^2 + y^2 - x^2y - xy - y = 0\)

=>\((x^3+x^2)-(x^2y+xy)+(y^2-y)=0\)

=>\(x^2(x+1)-xy(x+1)+y(y-1)=0\)

=>\(y^2 - (x^2 + x + 1)y + (x^3 + x^2) = 0\)

\(\Delta = (x^2 + x + 1)^2 - 4 \cdot 1 \cdot (x^3 + x^2)\)

\(=\left(x^2+x+1\right)^2-4\left(x^3+x^2\right)\)

\(=x^4+2x^3+3x^2+2x+1-4x^3-4x^2\)

\(=x^4-2x^3-x^2+2x+1=(x^2-x-1)^2\) >=0∀x

=>(1) có hai nghiệm là:

\(\left[\begin{array}{l}y=\frac{(x^2+x+1)+(x^2-x-1)}{2}=\frac{x^2+x+1+x^2-x-1}{2}=\frac{2x^2}{2}=x^2\\ y=\frac{(x^2+x+1)-(x^2-x-1)}{2}=\frac{x^2+x+1-x^2+x+1}{2}=\frac{2x+2}{2}=x+1\end{array}\right.\)

TH1: \(y = x^2\)

(2) sẽ trở thành: \(\sqrt{x} + \sqrt{x^2 - 1} = \sqrt{2x^2 - 3x - 4}\)

=>\(x + x^2 - 1 + 2\sqrt{x(x^2 - 1)} = 2x^2 - 3x - 4\)

=>\(2\sqrt{x^3 - x}=x^2-4x-3\)

=>\(\begin{cases}4(x^3-x)=(x^2-4x-3)^2\\ x^2-4x-3\ge0\end{cases}\)

=>\(\begin{cases}4x^3-4x=x^4+16x^2+9-8x^3-6x^2+24x\\ x^2-4x+4-7\ge0\end{cases}\)

=>\(\begin{cases}x^4-12x^3+10x^2+28x+9=0\\ \left(x-2\right)^2\ge7\end{cases}\Rightarrow\begin{cases}(x^2-10x-9)(x^2-2x-1)=0\\ \left(x-2\right)^2\ge7\end{cases}\)

=>\(x=5+\sqrt{34}\)

=>\(y=x^2=(5+\sqrt{34})^2=59+10\sqrt{34}\)

TH2: y=x+1

Thay y=x+1 vào (2), ta được:

\(\sqrt{x} + \sqrt{(x + 1) - 1} = \sqrt{2(x + 1) - 3x - 4}\)

=>\(\sqrt{x}+\sqrt{x}=\sqrt{2x + 2 - 3x - 4}\)

=>\(2\sqrt{x}=\sqrt{-x-2}\)

=>-x-2>=0 và 4x=-x-2

=>-x>=2 và 5x=-2

=>x<=-2 và x=-2/5

=>x∈∅

=>Loại

Bài 4:

\(x^4y-x^4+2x^3-2x^2+2x-y=1\)

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

\(\Leftrightarrow y(x^2+1)(x^2-1)-[x^2(x^2-2x+1)+(x^2-2x+1)]=0\)

\(\Leftrightarrow y(x^2+1)(x-1)(x+1)-(x-1)^2(x^2+1)=0\)

\(\Leftrightarrow (x^2+1)(x-1)[y(x+1)-(x-1)]=0\)

\(\Rightarrow \left[\begin{matrix} x-1=0(1)\\ y(x+1)-(x-1)=0(2)\end{matrix}\right.\)

Với $(1)$ ta thu được $x=1$, và mọi $ý$ nguyên.

Với $(2)$

\(y(x+1)=x-1\Rightarrow y=\frac{x-1}{x+1}\in\mathbb{Z}\)

\(\Rightarrow x-1\vdots x+1\)

\(\Rightarrow x+1-2\vdots x+1\Rightarrow 2\vdots x+1\)

\(\Rightarrow x+1\in\left\{\pm 1; \pm 2\right\}\Rightarrow x\in\left\{-2; 0; -3; 1\right\}\)

\(\Rightarrow y\left\{3;-1; 2; 0\right\}\)

Vậy \((x,y)=(-2,3); (0; -1); (-3; 2); (1; t)\) với $t$ nào đó nguyên.

Bài 1:

\(x^2+y^2-8x+3y=-18\)

\(\Leftrightarrow x^2+y^2-8x+3y+18=0\)

\(\Leftrightarrow (x^2-8x+16)+(y^2+3y+\frac{9}{4})=\frac{1}{4}\)

\(\Leftrightarrow (x-4)^2+(y+\frac{3}{2})^2=\frac{1}{4}\)

\(\Rightarrow (x-4)^2=\frac{1}{4}-(y+\frac{3}{2})^2\leq \frac{1}{4}<1\)

\(\Rightarrow -1< x-4< 1\Rightarrow 3< x< 5\)

Vì \(x\in\mathbb{Z}\Rightarrow x=4\)

Thay vào pt ban đầu ta thu được \(y=-1\) or \(y=-2\)

Vậy.......

a. Trừ vế theo vế \(\left(1\right)\) cho \(\left(2\right)\) ta được \(x^2-y^2=4x-4y\)

\(\Leftrightarrow\left(x-y\right)\left(x+y-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=4-y\end{matrix}\right.\)

TH1: \(x=y\)

Phương trình \(\left(1\right)\) tương đương:

\(x^2=2x\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=y=0\\x=y=2\end{matrix}\right.\)

TH2: \(x=4-y\)

Phương trình \(\left(2\right)\) tương đương:

\(y^2=4y-4\)

\(\Leftrightarrow y^2-4y+4=0\)

\(\Leftrightarrow\left(y-2\right)^2=0\)

\(\Leftrightarrow y=2\)

\(\Rightarrow x=2\)

Vậy hệ đã cho có nghiệm \(\left(x;y\right)\in\left\{\left(0;0\right);\left(2;2\right)\right\}\)

b. \(\left\{{}\begin{matrix}x+y+xy=5\\x^2+y^2=5\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}xy=5-\left(x+y\right)\\\left(x+y\right)^2-10+2\left(x+y\right)=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy=5-\left(x+y\right)\\\left(x+y\right)^2+2\left(x+y\right)-15=0\end{matrix}\right.\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+y=-5\\xy=10\end{matrix}\right.\\\left\{{}\begin{matrix}x+y=3\\xy=2\end{matrix}\right.\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}x+y=-5\\xy=10\end{matrix}\right.\Leftrightarrow\) vô nghiệm

TH2: \(\left\{{}\begin{matrix}x+y=3\\xy=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\\\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\end{matrix}\right.\)

Vậy ...

a: x>2

y>2

=>x+y>2+2=4

x>y>2

=>xy>2^2=4

b: x^2-xy=x(x-y)

x-y>0; x>0

=>x(x-y)>0

=>x^2-xy>0

y>2

=>y-2>0

=>y(y-2)>0

=>y^2-2y>0

x>y và y>2

=>y>0 và x-y>0

=>y(x-y)>0

=>xy-y^2>0