a) \(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne-3\\x\ne3\end{cases}}\)

\(A=\left(\frac{1}{3}+\frac{3}{x^2-3x}\right):\left(\frac{x^2}{27-3x^2}+\frac{1}{x+3}\right)\)\(=\left[\frac{1}{3}+\frac{3}{x\left(x-3\right)}\right]:\left(\frac{-x^2}{3x^2-27}+\frac{1}{x+3}\right)\)

\(=\left[\frac{x\left(x-3\right)}{3x\left(x-3\right)}+\frac{9}{3x\left(x-3\right)}\right]:\left[\frac{-x^2}{3\left(x^2-9\right)}+\frac{1}{x+3}\right]\)

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

\(=\frac{x^2-3x+9}{3x\left(x-3\right)}:\frac{-x^2+3x-9}{3\left(x-3\right)\left(x+3\right)}\)\(=\frac{x^2-3x+9}{3x\left(x-3\right)}.\frac{3\left(x-3\right)\left(x+3\right)}{-\left(x^2-3x+9\right)}=\frac{x+3}{-x}=\frac{-x-3}{x}=-1-\frac{3}{x}\)

b) \(A< -1\)\(\Leftrightarrow-1-\frac{3}{x}< -1\)\(\Leftrightarrow\frac{-3}{x}< 0\)

mà \(-3< 0\)\(\Rightarrow x>0\)và \(x\ne3\)

Vậy \(A< -1\Leftrightarrow\hept{\begin{cases}x>0\\x\ne3\end{cases}}\)

c) Vì \(-1\inℤ\)\(\Rightarrow\)Để A nguyên thì \(\frac{3}{x}\inℤ\)\(\Rightarrow3⋮x\)

\(\Rightarrow x\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

So sánh với ĐKXĐ \(\Rightarrow x=\pm3\)loại

Vậy A nguyên \(\Leftrightarrow x=\pm1\)

a)

ĐKXĐ: \(x\notin\left\{0;3;-3\right\}\)

Ta có: \(A=\left(\dfrac{1}{3}+\dfrac{3}{x^2-3x}\right):\left(\dfrac{x^2}{27-3x^2}+\dfrac{1}{x+3}\right)\)

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

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

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

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

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

b) Để A nguyên thì \(-x-3⋮x\)

mà \(-x⋮x\)

nên \(-3⋮x\)

\(\Leftrightarrow x\inƯ\left(-3\right)\)

\(\Leftrightarrow x\in\left\{1;-1;3;-3\right\}\)

Kết hợp ĐKXĐ, ta được: \(x\in\left\{1;-1\right\}\)

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

ai giup mink vs

\(\left(\frac{x^2+3x}{x^3+3x^2+9x+27}+\frac{3}{x^2+9}\right):\left(\frac{1}{x-3}-\frac{6x}{x^3-3x^2+9x-27}\right)\)

\(=\left(\frac{x\left(x+3\right)}{\left(x+3\right)\left(x^2+9\right)}+\frac{3}{x^2+9}\right):\left(\frac{1}{x-3}-\frac{6x}{\left(x-3\right)\left(x^2+9\right)}\right)\)

\(=\left(\frac{x}{x^2+9}+\frac{3}{x^2+9}\right):\left(\frac{x^2+9-6x}{\left(x-3\right)\left(x^2+9\right)}\right)=\frac{x+3}{x^2+9}:\frac{x^2+9-6x}{\left(x-3\right)\left(x^2+9\right)}\)

\(=\frac{\left(x+3\right)\left(x-3\right)\left(x^2+9\right)}{\left(x^2+9\right)\left(x^2-6x+9\right)}=\frac{\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x-3\right)}=\frac{x+3}{x-3}\)

b) \(Voix>0\Rightarrow P\ne\varnothing\)(mk ko chac)

c) \(P\inℤ\Leftrightarrow x+3⋮x-3\Leftrightarrow x-3\in\left\{-1;-2;-3;-6;1;2;3;6\right\}\) 

sau do tinh

cau nay la toan lp 8 nha

P= O/ nha

a: \(A=\left(\dfrac{x}{x^2-4}+\dfrac{4}{x-2}+\dfrac{1}{x+2}\right):\dfrac{3x+3}{x^2+2x}\)

\(=\dfrac{x+4x+8+x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x+2\right)}{3\left(x+1\right)}\)

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

\(=\dfrac{2x}{x-2}\)

Không chép lại đề nhé:

\(1A=\left(\frac{x\left(x+3\right)}{\left(x+3\right)\left(x^2+9\right)}+\frac{3}{x^2+9}\right):\left(\frac{1}{x-3}-\frac{6x}{\left(x-3\right)\left(x^2+9\right)}\right)\)

\(=\frac{x+3}{x^2+9}:\frac{x^2+9-6x}{\left(x-3\right)\left(x^2+9\right)}\)

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

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

b/ Với x > 0 thì P không xác định khi x = 3 (vì mẫu sẽ = 0)

c/ \(A=\frac{x+3}{x-3}=1+\frac{6}{x-3}\)

Để A nguyên thì (x - 3) phải là ước nguyên của 6 hay

(x - 3) \(\in\)(- 1; - 2; - 3, - 6; 1; 2; 3; 6)

Thế vào sẽ tìm được A

ĐKXĐ thì b tự làm nhé