a) Điều kiện xác định của biểu thức \(A\) là: \(x^{2} - 4 \neq 0 ; x - 2 \neq 0\) và \(x + 2 \neq 0\)

Mà \(x^{2} - 4 = \left(\right. x - 2 \left.\right) \left(\right. x + 2 \left.\right)\)

Vậy điều kiện xác định của biểu thức \(A\) là \(x - 2 \neq 0\) và \(x + 2 \neq 0\) hay \(x \neq \&\text{nbsp}; 2\) và \(x \neq - 2\).

b) Với điều kiện xác định \(x \neq \&\text{nbsp}; 2\) và \(x \neq - 2\) ta có:

\(A = \frac{2 x^{2}}{x^{2} - 4} - \frac{x}{x - 2} - \frac{2}{x + 2}\)

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

\(= \frac{2 x^{2} - x^{2} - 2 x - 2 x + 4}{\left(\right. x - 2 \left.\right) \left(\right. x + 2 \left.\right)}\)

\(= \frac{x^{2} - 4 x + 4}{\left(\right. x - 2 \left.\right) \left(\right. x + 2 \left.\right)}\)

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

\(= \frac{x - 2}{x + 2} .\)

c) Với \(x \neq \&\text{nbsp}; 2 ,\) và \(x \neq - 2\) để \(A = 2\) thì \(\frac{x - 2}{x + 2} = 2\)

Suy ra \(x - 2 = 2 \left(\right. x + 2 \left.\right)\)

Do đó \(x - 2 = 2 x + 4\) hay \(x = - 6\) (thỏa mãn điều kiện)

Vậy \(x = - 6.\)

a) \(x y + y^{2} - x - y\)

\(= \left(\right. x y + y^{2} \left.\right) - \left(\right. x + y \left.\right)\)

\(= y \left(\right. x + y \left.\right) - \left(\right. x + y \left.\right)\)

\(= \left(\right. x + y \left.\right) \left(\right. y - 1 \left.\right) .\)​

b) \(\left(\left(\right. x^{2} y^{2} - 8 \left.\right)\right)^{2} - 1\)

\(= \left(\right. x^{2} y^{2} - 8 - 1 \left.\right) \left(\right. x^{2} y^{2} - 8 + 1 \left.\right)\)

\(= \left(\right. x^{2} y^{2} - 9 \left.\right) \left(\right. x^{2} y^{2} - 7 \left.\right)\)

\(= \left(\right. x y - 3 \left.\right) \left(\right. x y + 3 \left.\right) \left(\right. x^{2} y^{2} - 7 \left.\right) .\)

\(= \left(\right. x - 1 \left.\right) \left(\right. x + 8 \left.\right) .\)

a) \(x y + y^{2} - x - y\)

\(= \left(\right. x y + y^{2} \left.\right) - \left(\right. x + y \left.\right)\)

\(= y \left(\right. x + y \left.\right) - \left(\right. x + y \left.\right)\)

\(= \left(\right. x + y \left.\right) \left(\right. y - 1 \left.\right) .\)​

b) \(\left(\left(\right. x^{2} y^{2} - 8 \left.\right)\right)^{2} - 1\)

\(= \left(\right. x^{2} y^{2} - 8 - 1 \left.\right) \left(\right. x^{2} y^{2} - 8 + 1 \left.\right)\)

\(= \left(\right. x^{2} y^{2} - 9 \left.\right) \left(\right. x^{2} y^{2} - 7 \left.\right)\)

\(= \left(\right. x y - 3 \left.\right) \left(\right. x y + 3 \left.\right) \left(\right. x^{2} y^{2} - 7 \left.\right) .\)

\(= \left(\right. x - 1 \left.\right) \left(\right. x + 8 \left.\right) .\)

a) \(\left(\right. - 12 x^{13} y^{15} + 6 x^{10} y^{14} \left.\right) : \left(\right. - 3 x^{10} y^{14} \left.\right)\)

\(= \left(\right. - 12 x^{13} y^{15} \left.\right) : \left(\right. - 3 x^{10} y^{14} \left.\right) + \left(\right. 6 x^{10} y^{14} \left.\right) : \left(\right. - 3 x^{10} y^{14} \left.\right)\)

\(= 4 x^{3} y - 2.\)

b) \(\left(\right. x - y \left.\right) \left(\right. x^{2} - 2 x + y \left.\right) - x^{3} + x^{2} y\)

\(= x \left(\right. x^{2} - 2 x + y \left.\right) - y \left(\right. x^{2} - 2 x + y \left.\right) - x^{3} + x^{2} y\)

\(= x^{3} - 2 x^{2} + x y - x^{2} y + 2 x y - y^{2} - x^{3} + x^{2} y\)

\(= - 2 x^{2} + 3 x y - y^{2} .\)

\(a)1s^22s^22p^{^6}3s^23p^6\)

\(ô=20,chukì3,nhómVIIA\)