Giải hệ phương trình

\(\left\{{}\begin{matrix}\dfrac{x}{y}-\dfrac{x}{y+12}=1\\\dfrac{x}{y-12}-\dfrac{x}{y}=2\end{matrix}\right.\)

Write two notices about your class meeting and your Sports Club meeting.

[Unit 7- SGK]

[Please help me! Thanks!]

\(\frac{x\left(x-3\right)-2x+6}{x^2-x}=\frac{x^2-3x-2x+6}{x\left(x-2\right)}=\frac{\left(x^2-2x\right)-\left(3x-6\right)}{x\left(x-2\right)}\)

\(=\frac{x\left(x-2\right)-3\left(x-2\right)}{x\left(x-2\right)}=\frac{\left(x-3\right)\left(x-2\right)}{x\left(x-2\right)}=\frac{x-3}{x}\)

Tìm các số nguyên x, y thỏa mãn: (x² + y² +1)² -5x² - 4y² - 5 = 0

4. \(\left(x-1\right)^3-\left(x+1\right)^3+6\left(x+1\right)\left(x-1\right)\)

\(=\left(x^3-3x^2+3x-1\right)-\left(x^3+3x^2+3x+1\right)+\left(6x+1\right)\left(x-1\right)\)\(=x^3-3x^2+3x-1-x^3-3x^2-3x-1+6x^2-6\)

\(=\left(x^3-x^3\right)+\left(-3x^2-3x^2\right)+6x^2+\left(3x-3x\right)-1-1-6\)\(=-8\)

3. \(\left(x+2\right)^3+\left(x-2\right)^3-2x\left(x^2+12\right)\)

\(=x^3+3x^22+3x2^2+2^3+x^3-3x^22+3x2^2-2^3-2x^3-24x\)

\(=x^3+6x^2+12x+8+x^3-6x^2+12x-8-2x^3-24x\)

\(=\left(x^3+x^3\right)-2x^3+\left(6x^2-6x^2\right)+\left(12x+12x\right)-24x+\left(8-8\right)\)\(=0\)