a) Thay \(x = 2\) (thỏa mãn điều kiện xác định) vào \(Q = \frac{x + 1}{x^{2} - 9}\), ta được:

\(Q = \frac{x + 1}{x^{2} - 9} = \frac{2 + 1}{2^{2} - 9} = \frac{3}{- 5} = - \frac{3}{5}\)

b) \(P = \frac{2 x^{2} - 1}{x \left(\right. x + 1 \left.\right)} - \frac{\left(\right. x - 1 \left.\right) \left(\right. x + 1 \left.\right)}{x \left(\right. x + 1 \left.\right)} + \frac{3 x}{x \left(\right. x + 1 \left.\right)}\)

\(P = \frac{2 x^{2} - 1 - \left(\right. x^{2} - 1 \left.\right) + 3 x}{x \left(\right. x + 1 \left.\right)}\)

\(P = \frac{2 x^{2} - 1 - x^{2} + 1 + 3 x}{x \left(\right. x + 1 \left.\right)}\)

\(P = \frac{x^{2} + 3 x}{x \left(\right. x + 1 \left.\right)} = \frac{x + 3}{x + 1}\)

c) Ta có \(M = P . Q = \frac{x + 3}{x + 1} . \frac{x + 1}{x^{2} - 9} = \frac{x + 3}{\left(\right. x - 3 \left.\right) \left(\right. x + 3 \left.\right)} = \frac{1}{x - 3}\)

\(M = \frac{- 1}{2}\) suy ra \(\frac{1}{x - 3} = \frac{- 1}{2}\)

\(x - 3 = - 2\)

\(x = 1\).

Vậy với \(x = 1\) thì\(M = \frac{- 1}{2}\)

a) Thay \(x = 2\) (thỏa mãn điều kiện xác định) vào \(Q = \frac{x + 1}{x^{2} - 9}\), ta được:

\(Q = \frac{x + 1}{x^{2} - 9} = \frac{2 + 1}{2^{2} - 9} = \frac{3}{- 5} = - \frac{3}{5}\)

b) \(P = \frac{2 x^{2} - 1}{x \left(\right. x + 1 \left.\right)} - \frac{\left(\right. x - 1 \left.\right) \left(\right. x + 1 \left.\right)}{x \left(\right. x + 1 \left.\right)} + \frac{3 x}{x \left(\right. x + 1 \left.\right)}\)

\(P = \frac{2 x^{2} - 1 - \left(\right. x^{2} - 1 \left.\right) + 3 x}{x \left(\right. x + 1 \left.\right)}\)

\(P = \frac{2 x^{2} - 1 - x^{2} + 1 + 3 x}{x \left(\right. x + 1 \left.\right)}\)

\(P = \frac{x^{2} + 3 x}{x \left(\right. x + 1 \left.\right)} = \frac{x + 3}{x + 1}\)

c) Ta có \(M = P . Q = \frac{x + 3}{x + 1} . \frac{x + 1}{x^{2} - 9} = \frac{x + 3}{\left(\right. x - 3 \left.\right) \left(\right. x + 3 \left.\right)} = \frac{1}{x - 3}\)

\(M = \frac{- 1}{2}\) suy ra \(\frac{1}{x - 3} = \frac{- 1}{2}\)

\(x - 3 = - 2\)

\(x = 1\).

Vậy với \(x = 1\) thì\(M = \frac{- 1}{2}\)

a) Thay \(x = 2\) (thỏa mãn điều kiện xác định) vào \(Q = \frac{x + 1}{x^{2} - 9}\), ta được:

\(Q = \frac{x + 1}{x^{2} - 9} = \frac{2 + 1}{2^{2} - 9} = \frac{3}{- 5} = - \frac{3}{5}\).

b) \(P = \frac{2 x^{2} - 1}{x \left(\right. x + 1 \left.\right)} - \frac{\left(\right. x - 1 \left.\right) \left(\right. x + 1 \left.\right)}{x \left(\right. x + 1 \left.\right)} + \frac{3 x}{x \left(\right. x + 1 \left.\right)}\)

\(P = \frac{2 x^{2} - 1 - \left(\right. x^{2} - 1 \left.\right) + 3 x}{x \left(\right. x + 1 \left.\right)}\)

\(P = \frac{2 x^{2} - 1 - x^{2} + 1 + 3 x}{x \left(\right. x + 1 \left.\right)}\)

\(P = \frac{x^{2} + 3 x}{x \left(\right. x + 1 \left.\right)} = \frac{x + 3}{x + 1}\)

c) Ta có \(M = P . Q = \frac{x + 3}{x + 1} . \frac{x + 1}{x^{2} - 9} = \frac{x + 3}{\left(\right. x - 3 \left.\right) \left(\right. x + 3 \left.\right)} = \frac{1}{x - 3}\)

\(M = \frac{- 1}{2}\) suy ra \(\frac{1}{x - 3} = \frac{- 1}{2}\)

\(x - 3 = - 2\)

\(x = 1\).

Vậy với \(x = 1\) thì\(M = \frac{- 1}{2}\)

a) Thay \(x = 2\) (thỏa mãn điều kiện xác định) vào \(Q = \frac{x + 1}{x^{2} - 9}\), ta được:

\(Q = \frac{x + 1}{x^{2} - 9} = \frac{2 + 1}{2^{2} - 9} = \frac{3}{- 5} = - \frac{3}{5}\).

b) \(P = \frac{2 x^{2} - 1}{x \left(\right. x + 1 \left.\right)} - \frac{\left(\right. x - 1 \left.\right) \left(\right. x + 1 \left.\right)}{x \left(\right. x + 1 \left.\right)} + \frac{3 x}{x \left(\right. x + 1 \left.\right)}\)

\(P = \frac{2 x^{2} - 1 - \left(\right. x^{2} - 1 \left.\right) + 3 x}{x \left(\right. x + 1 \left.\right)}\)

\(P = \frac{2 x^{2} - 1 - x^{2} + 1 + 3 x}{x \left(\right. x + 1 \left.\right)}\)

\(P = \frac{x^{2} + 3 x}{x \left(\right. x + 1 \left.\right)} = \frac{x + 3}{x + 1}\)

c) Ta có \(M = P . Q = \frac{x + 3}{x + 1} . \frac{x + 1}{x^{2} - 9} = \frac{x + 3}{\left(\right. x - 3 \left.\right) \left(\right. x + 3 \left.\right)} = \frac{1}{x - 3}\)

\(M = \frac{- 1}{2}\) suy ra \(\frac{1}{x - 3} = \frac{- 1}{2}\)

\(x - 3 = - 2\)

\(x = 1\).

Vậy với \(x = 1\) thì\(M = \frac{- 1}{2}\)

a) Thay \(x = 2\) (thỏa mãn điều kiện xác định) vào \(Q = \frac{x + 1}{x^{2} - 9}\), ta được:

\(Q = \frac{x + 1}{x^{2} - 9} = \frac{2 + 1}{2^{2} - 9} = \frac{3}{- 5} = - \frac{3}{5}\).

b) \(P = \frac{2 x^{2} - 1}{x \left(\right. x + 1 \left.\right)} - \frac{\left(\right. x - 1 \left.\right) \left(\right. x + 1 \left.\right)}{x \left(\right. x + 1 \left.\right)} + \frac{3 x}{x \left(\right. x + 1 \left.\right)}\)

\(P = \frac{2 x^{2} - 1 - \left(\right. x^{2} - 1 \left.\right) + 3 x}{x \left(\right. x + 1 \left.\right)}\)

\(P = \frac{2 x^{2} - 1 - x^{2} + 1 + 3 x}{x \left(\right. x + 1 \left.\right)}\)

\(P = \frac{x^{2} + 3 x}{x \left(\right. x + 1 \left.\right)} = \frac{x + 3}{x + 1}\)

c) Ta có \(M = P . Q = \frac{x + 3}{x + 1} . \frac{x + 1}{x^{2} - 9} = \frac{x + 3}{\left(\right. x - 3 \left.\right) \left(\right. x + 3 \left.\right)} = \frac{1}{x - 3}\)

\(M = \frac{- 1}{2}\) suy ra \(\frac{1}{x - 3} = \frac{- 1}{2}\)

\(x - 3 = - 2\)

\(x = 1\).

Vậy với \(x = 1\) thì\(M = \frac{- 1}{2}\)

1. My favorite hobby is playing badminton

2. I stared playing when I was 11 years old

3.I often play with friends

4.We often play at the cultural center

5. I think playing badminton is a great way to stay fit and have fun