a) \(ĐKXĐ:\hept{\begin{cases}x\ne2\\x\ne3\end{cases}}\)

\(A=\frac{2x-9}{x^2-5x+6}-\frac{x+3}{x-2}-\frac{2x+4}{3-x}\)

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

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

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

\(\Leftrightarrow A=\frac{x^2+2x-8}{\left(x-2\right)\left(x-3\right)}\)

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

\(\Leftrightarrow A=\frac{x+4}{x-3}\)

b) Để \(A\inℤ\)

\(\Leftrightarrow\frac{x+4}{x-3}\inℤ\)

\(\Leftrightarrow1+\frac{7}{x-3}\inℤ\)

\(\Leftrightarrow x-3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

\(\Leftrightarrow x\in\left\{2;4;-4;10\right\}\)

Vậy để \(A\inℤ\Leftrightarrow x\in\left\{2;4;-4;10\right\}\)

c) Để \(A=\frac{3}{5}\)

\(\Leftrightarrow\frac{x+4}{x-3}=\frac{3}{5}\)

\(\Leftrightarrow5x+20=3x-9\)

\(\Leftrightarrow2x+29=0\)

\(\Leftrightarrow x=-\frac{29}{2}\)

d) Để \(A< 0\)

\(\Leftrightarrow\frac{x+4}{x-3}< 0\)

\(\Leftrightarrow1+\frac{7}{x-3}< 0\)

\(\Leftrightarrow\frac{-7}{x-3}< 1\)

\(\Leftrightarrow-7< x-3\)

\(\Leftrightarrow x>-4\)

e) Để \(A>0\)

\(\Leftrightarrow\frac{x+4}{x-3}>0\)

\(\Leftrightarrow1+\frac{7}{x-3}>0\)

\(\Leftrightarrow\frac{-7}{x-3}>1\)

\(\Leftrightarrow-7>x-3\)

\(\Leftrightarrow x< -4\)

sau khi rút gọn ta được \(P=\frac{x-4}{x-2}\left(x\ne-3;x\ne2;x\ne-2\right)\)

d,ta có \(P=\frac{x-4}{x-2}=\frac{x-2-2}{x-2}=1-\frac{2}{x-2}\left(x\ne-2;x\ne-3;x\ne2\right)\)

để P nguyên mà x nguyên \(\Leftrightarrow x-2\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)

ta có bảng:

vậy \(P\in Z\Leftrightarrow x\in\left\{3;1;4;0\right\}\)

e,x²-9=0

\(\Leftrightarrow x^2=9\Leftrightarrow\orbr{\begin{cases}x=3\left(tm\right)\\x=-3\left(kotm\right)\end{cases}}\)

thay x=3 vào P đã rút gọn ta có \(P=\frac{3-4}{3-2}=-1\)

vậy với x=3 thì p có giá trị bằng -1

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bạn viết thế này khó nhìn quá

nhìn hơi đau mắt nhá bạn hoa mắt quá

a,ĐK:  \(\hept{\begin{cases}x\ne0\\x\ne\pm3\end{cases}}\)

b, \(A=\left(\frac{9}{x\left(x-3\right)\left(x+3\right)}+\frac{1}{x+3}\right):\left(\frac{x-3}{x\left(x+3\right)}-\frac{x}{3\left(x+3\right)}\right)\)

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

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

c, Với x = 4 thỏa mãn ĐKXĐ thì

\(A=\frac{-3}{4-3}=-3\)

d, \(A\in Z\Rightarrow-3⋮\left(x-3\right)\)

\(\Rightarrow x-3\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\Rightarrow x\in\left\{0;2;4;6\right\}\)

Mà \(x\ne0\Rightarrow x\in\left\{2;4;6\right\}\)

a) đk: \(x\ne0;4\); \(x>0\)

P = \(\left[\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{1}{\sqrt{x}-2}\right]\times\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)\)

= \(\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\times\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)\)

= \(\dfrac{1}{\sqrt{x}\left(\sqrt{x}-2\right)}.\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)

b) Để P < \(\dfrac{1}{2}\)

<=> \(\dfrac{\sqrt{x}-1}{\sqrt{x}}< \dfrac{1}{2}\)

<=> \(1-\dfrac{1}{\sqrt{x}}< \dfrac{1}{2}\)

<=> \(\dfrac{1}{\sqrt{x}}>\dfrac{1}{2}\)

<=> \(\sqrt{x}< 2\)

<=> x < 4

<=> 0 < x < 4

c, Tìm gt nguyên của x để P có gt nguyên

a) ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\)

a) ĐKXĐ: x \(\ge\)0; x \(\ne\)4

Ta có: P = \(\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{x+5}{x-\sqrt{x}-2}\)

P = \(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}-\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}-\frac{x+5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)

P = \(\frac{x-3\sqrt{x}+2-x-4\sqrt{x}-3-x-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)

P = \(\frac{-x-7\sqrt{x}-6}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)

P = \(\frac{-\left(x+6\sqrt{x}+\sqrt{x}+6\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)

P = \(\frac{-\left(\sqrt{x}+1\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)

P = \(-\frac{\sqrt{x}+6}{\sqrt{x}-2}\)

b) Với x \(\ge\)0 và x \(\ne\)4, ta có:

P > -1 <=> \(-\frac{\sqrt{x}+6}{\sqrt{x}-2}>-1\)

<=> \(-\frac{\sqrt{x}+6}{\sqrt{x}-2}+1>0\)

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

<=> \(\frac{-8}{\sqrt{x}-2}>0\)

Do -8 < 0 => \(\sqrt{x}-2< 0\) <=> \(\sqrt{x}< 2\)<=> \(x< 4\)

mà x \(\ge0\) => 0 \(\le\)x \(< \)4

c)Với x \(\ge\)0 và x \(\ne\)4

Để P \(\in\)Z <=> -8 \(-8⋮\sqrt{x}-2\)

<=> \(\sqrt{x}-2\inƯ\left(-8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)

Do \(\sqrt{x}\ge0\) <=> \(\sqrt{x}-2\ge-2\) => \(\sqrt{x}-2\in\left\{-2;-1;1;2;4;8\right\}\)

Lập bảng: 

Vậy ....