\(-\left|1,7-x\right|-\dfrac{5}{3}=\dfrac{2}{3}\\ \Rightarrow\left|1,7-x\right|=-\dfrac{5}{3}-\dfrac{2}{3}=-\dfrac{7}{3}\left(l\right)\)

Vậy không có giá trị x thoả mãn

\(\left(x-3\right)\cdot\left(y-5\right)=3\)

=>\(\left(x-3\right)\cdot\left(y-5\right)=1\cdot3=3\cdot1=\left(-1\right)\cdot\left(-3\right)=\left(-3\right)\cdot\left(-1\right)\)

=>\(\left(x-3;y-5\right)\in\left\{\left(1;3\right);\left(3;1\right);\left(-1;-3\right);\left(-3;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(4;8\right);\left(6;6\right);\left(2;2\right);\left(0;4\right)\right\}\)

Ta có: \(x+y=7\Rightarrow\left(x+y\right)^2=49\Rightarrow x^2+y^2+2xy=49\)

Mà: \(x^2+y^2=25\Rightarrow2xy=24\Rightarrow xy=12\)

\(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=7\left(25-12\right)=91\)

(Vì\(x+y=7;x^2+y^2=25;xy=12\))

\(\dfrac{x+2}{x-3}+\dfrac{x-2}{x}=\dfrac{x^2+2x+6}{x\left(x-3\right)}\)  đkxđ: x khác 3 , x khác 0

\(\Leftrightarrow\dfrac{x\left(x+2\right)}{x\left(x-3\right)}+\dfrac{\left(x-2\right)\left(x-3\right)}{x\left(x-3\right)}-\dfrac{x^2+2x+6}{x\left(x-3\right)}=0\)

\(\Leftrightarrow\dfrac{x^2+2x}{....}+\dfrac{x^2-3x-2x+6}{.....}-\dfrac{x^2+2x+6}{...}=0\)

\(\Leftrightarrow x^2+2x+x^2-3x-2x+6-x^2-2x-6=0\)

\(\Leftrightarrow x^2-5x=0\)

\(\Leftrightarrow x\left(x-5\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}x=0\left(tm\right)\\x=5\left(tm\right)\end{matrix}\right.\)

đk : x khác -3 ; 3

\(\Rightarrow\left(x+3\right)^2-4x^2=\left(x-3\right)^2\Leftrightarrow-3x^2+6x+9=x^2-6x+9\)

\(\Leftrightarrow4x^2-12x=0\Leftrightarrow4x\left(x-3\right)=0\Leftrightarrow x=3\left(ktm\right);x=0\)

\(\dfrac{x+3}{x-3}-\dfrac{4x^2}{x^2-9}=\dfrac{x-3}{x+3}\) đkxđ: x khác 3 ; x khác -3

\(\Leftrightarrow\dfrac{x+3}{x-3}-\dfrac{4x^2}{\left(x-3\right)\left(x+3\right)}-\dfrac{x-3}{x+3}=0\)

\(\Leftrightarrow\dfrac{\left(x+3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{4x^2}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(x-3\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=0\)

\(\Leftrightarrow\dfrac{x^2+6x+9}{...}-\dfrac{4x^2}{.....}-\dfrac{x^2-6x+9}{....}=0\)

\(\Leftrightarrow x^2+6x+9-4x^2-x^2+6x-9=0\)

\(\Leftrightarrow-4x^2+12x=0\)

\(\Leftrightarrow-4x\left(x-3\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\left(tm\right)\\x=3\left(kotm\right)\end{matrix}\right.\)

Vậy tập nghiệm của pt có S thuộc \(\left\{0\right\}\)

x, y có điều kiện gì không bạn?

ko

\(xy=x-y+3\)

\(\Leftrightarrow xy-x+y=3\)

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

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

\(\Leftrightarrow x+1;y-1\inƯ\left(2\right)\)

Ta có: \(Ư\left(2\right)=\left\{\pm1;\pm2\right\}\)

Lập bảng:

Vậy các cặp số nguyên (x; y) thỏa mãn là (-2; -1); (0;3); (-3; 0) và (1; 2)