Bài 1: 

a: \(A=\dfrac{x^2-3+x+3}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x}=\dfrac{x\left(x+1\right)}{x\left(x-3\right)}=\dfrac{x+1}{x-3}\)

b: Để A=3 thì 3x-9=x+1

=>2x=10

hay x=5

Bài 2: 

a: \(A=\dfrac{x+x-2-2x-4}{\left(x-2\right)\left(x+2\right)}:\dfrac{x+2-x}{x+2}\)

\(=\dfrac{-6}{x-2}\cdot\dfrac{1}{2}=\dfrac{-3}{x-2}\)

b: Để A nguyên thì \(x-2\in\left\{1;-1;3;-3\right\}\)

hay \(x\in\left\{3;1;5;-1\right\}\)

a) \(A=\frac{2x}{x+3}-\frac{x+1}{3-x}-\frac{3-11x}{x^2-9}\)

\(\Leftrightarrow A=\frac{2x}{x+3}+\frac{x+1}{x-3}-\frac{3-11x}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow A=\frac{2x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{3-11x}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow A=\frac{2x^2-6x}{\left(x+3\right)\left(x-3\right)}+\frac{x^2+4x+3}{\left(x-3\right)\left(x+3\right)}-\frac{3-11x}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow A=\frac{3x^2-13x}{x^2-9}\)

\(A=\frac{2x}{x+3}-\frac{x+1}{3-x}-\frac{3-11x}{x^2-9}\)

a) ĐK : x ≠ ±3

\(=\frac{2x}{x+3}+\frac{x+1}{x-3}-\frac{3-11x}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{2x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{3-11x}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{2x^2-6x}{\left(x-3\right)\left(x+3\right)}+\frac{x^2+4x+3}{\left(x-3\right)\left(x+3\right)}-\frac{3-11x}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{2x^2-6x+x^2+4x+3-3+11x}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{3x^2+9x}{\left(x-3\right)\left(x+3\right)}\)

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

b) Để A < 2

=> \(\frac{3x}{x-3}< 2\)

<=> \(\frac{3x}{x-3}-2< 0\)

<=> \(\frac{3x}{x-3}-\frac{2x-6}{x-3}< 0\)

<=> \(\frac{3x-2x+6}{x-3}< 0\)

<=> \(\frac{x+6}{x-3}< 0\)

Xét hai trường hợp :

1. \(\hept{\begin{cases}x+6>0\\x-3< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>-6\\x< 3\end{cases}}\Leftrightarrow-6< x< 3\)

2. \(\hept{\begin{cases}x+6< 0\\x-3>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x< -6\\x>3\end{cases}}\)( loại )

Vậy -6 < x < 3

dell hiểu đc đề ntn cả?

ngu nên đéo hiểu

trả lời hộ mình tí

1.

\(A=\dfrac{2x-9}{\left(x-2\right)\left(x-3\right)}-\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-2\right)\left(x-3\right)}+\dfrac{\left(2x+4\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)

\(=\dfrac{2x-9-\left(x^2-9\right)+\left(2x^2-8\right)}{\left(x-2\right)\left(x-3\right)}\)

\(=\dfrac{x^2+2x-8}{\left(x-2\right)\left(x-3\right)}=\dfrac{\left(x-2\right)\left(x+4\right)}{\left(x-2\right)\left(x-3\right)}\)

\(=\dfrac{x+4}{x-3}\)

b.

\(A=2\Rightarrow\dfrac{x+4}{x-3}=2\Rightarrow x+4=2\left(x-3\right)\)

\(\Rightarrow x=10\) (thỏa mãn)

2.

\(x^4+2x^2y+y^2-9=\left(x^2+y\right)^2-3^2=\left(x^2+y-3\right)\left(x^2+y+3\right)\)

a: Thay x=-4 vào B, ta được:

\(B=\dfrac{-4+3}{-4}=\dfrac{-1}{-4}=\dfrac{1}{4}\)

b: \(P=A\cdot B=\dfrac{x^2-3x+2x-9+3x+9}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x}\)

\(=\dfrac{x^2+2x}{\left(x-3\right)}\cdot\dfrac{1}{x}=\dfrac{x+2}{x-3}\)

c: Để P nguyên thì \(x-3\in\left\{1;-1;5;-5\right\}\)

hay \(x\in\left\{4;2;8;-2\right\}\)

1) A = \(\dfrac{2x-1}{x+3}\) = \(\dfrac{3}{2}\) (=) (2x-1).2 = 3.(x+3)

                          (=) 4x-2 =3x+9

                          (=) 4x-3x = 9+2

                         (=) x = 11 (tm)

2) Để \(\dfrac{A}{B}\)< \(^{x^2}\)+5 (=) \(\dfrac{2x-1}{x+3}\): \(\dfrac{2}{x^2-9}\) <  \(x^2\)+5 

                    (=) \(\dfrac{\left(2x-1\right)}{\left(x+3\right)}.\dfrac{\left(x-3\right)\left(x+3\right)}{2}\) < \(x^2\)+5

                    (=) \(\dfrac{\left(2x-1\right).\left(x-3\right)}{2}< x^2+5\)

                    (=) \(\dfrac{2x^2-6x-x+3}{2}\) < \(x^2\) +5

                    (=) \(2x^2\)- 7x + 3 < \(2x^2\)+ 10

                    (=)  (\(2x^2\)-\(2x^2\)) - 7x < -3 +10

                    (=) -7x < 7 

                    (=) x > -1

Bạn nên gõ đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu đề và hỗ trợ bạn tốt hơn nhé.