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

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

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

Ta thấy   x² + x - 2  <   x² + x + 2

nên M < 1

a)  \(ĐKXĐ:x\ne\pm2\)

\(D=\frac{3x}{x-2}+\frac{2}{x+2}-\frac{14x-4}{x^2-4}:\frac{x\left(x-1\right)}{x+2}\)

\(\Leftrightarrow D=\frac{3x^2+6x+2x-4-14x+4}{\left(x-2\right)\left(x+2\right)}\cdot\frac{x+2}{x\left(x-1\right)}\)

\(\Leftrightarrow D=\frac{3x^2-6x}{x\left(x-1\right)\left(x-2\right)}\)

\(\Leftrightarrow D=\frac{3x\left(x-2\right)}{x\left(x-1\right)\left(x-2\right)}\)

\(\Leftrightarrow D=\frac{3}{x-1}\)

b) Khi \(\left|x-1\right|-3=0\)

\(\Leftrightarrow\left|x-1\right|=3\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=3\\1-x=3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=4\left(tm\right)\\x=-2\left(ktm\right)\end{cases}}\)

Thay \(x=4\)vào D ta được :\(D=\frac{3}{4-1}=1\)

c) Để D có giá trị nguyên

\(\Leftrightarrow\frac{3}{x-1}\)có giá trị nguyên

\(\Leftrightarrow x-1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

\(\Leftrightarrow x\in\left\{0;2;-2;4\right\}\)

Loại bỏ giá trị \(x=\pm2\)không làm cho biểu thức có nghĩa

Vậy để D có giá trị nguyên \(\Leftrightarrow x\in\left\{0;4\right\}\)

Khi làm bài thì chỉnh lại giúp bạn cái đề: 

\(D=\left(\frac{3X}{X-2}+\frac{2}{X+2}-\frac{14X-4}{X^2-4}\right):\frac{X\left(X-1\right)}{X+2}\)

\(A=\frac{1-x^2}{x}.\left(\frac{x^2}{x+3}-1\right)+\frac{3x^2-14x+3}{x^2+3x}\)

\(A=\frac{\left(x^2-x-3\right)\left(-x^2+1\right)}{x\left(x+3\right)}+\frac{3x^2-14x+3}{x^2+3x}\)

\(A=\frac{\left(x^2-x-3\right)\left(1-x^3\right)}{\left(x+3\right)x}+\frac{3x^2-14x+3}{x\left(x+3\right)}\)

\(A=\frac{\left(x^2-x-3\right)\left(1-x^2\right)+3x^2-14x+3}{\left(x+3\right)x}\)

\(A=\frac{-x^4+x^3+7x^2-15x}{x\left(x+3\right)}\)

\(A=\frac{x\left(-x^3+x^2+7x-15\right)}{x\left(x+3\right)}\)

\(A=\frac{-x^3+x^2+7x-15}{x+3}\)

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

\(A=-\left(x^2-4x+5\right)\)

\(A=-x^2+4x-5\)

Trình độ hơi thấp, có gì sai sót xin bỏ qua cho :)

umk cảm ơn bạn trước nhé

Phân thức A xác định \(\Leftrightarrow\hept{\begin{cases}x\ne0\\x+3\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-3\end{cases}}}\)

\(A=\frac{1-x^2}{x}\left(\frac{x^2}{x+3}-1\right)+\frac{3x^2-14x+3}{x^2+3x}\)

\(A=\frac{\left(1-x\right)\left(1+x\right)}{x}\cdot\frac{x^2-x-3}{x+3}+\frac{3x^2-14x+3}{x\left(x+3\right)}\)

\(A=\frac{\left(1-x\right)\left(1+x\right)\left(x^2-x-3\right)}{x\left(x+3\right)}+\frac{3x^2-14x+3}{x\left(x+3\right)}\)

\(A=\frac{-x^4+x^3+4x^2-x-3+3x^2-14x+3}{x\left(x+3\right)}\)

\(A=\frac{-x^4+x^3+7x^2-15x}{x\left(x+3\right)}\)

\(A=\frac{x\left(-x^3+x^2+7x-15\right)}{x\left(x+3\right)}\)

\(A=\frac{-x^3+x^2+7x-15}{x+3}\)

Phân tích tử thức : 

\(-x^3+x^2+7x-15\)

\(=-x^3-3x^2+4x^2+12x-5x-15\)

\(=-x^2\left(x+3\right)+4x\left(x+3\right)-5\left(x+3\right)\)

\(=\left(x+3\right)\left(-x^2+4x-5\right)\)

\(\Rightarrow A=\frac{\left(x+3\right)\left(-x^2+4x-5\right)}{x+3}=-x^2+4x-5\)

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 \(\Leftrightarrow x=\pm1\)