a) A có nghĩa \(\Leftrightarrow\left(x+1\right)^2-3x\ne0\), \(x^3+1\ne0\),\(x+1\ne0\),\(3x^2+6x\ne0\) và \(x^2-4\ne0\)

+) \(\left(x+1\right)^2-3x\ne0\Leftrightarrow x^2+2x+1-3x\ne0\)

\(\Leftrightarrow x^2-x+1\ne0\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ne0\)(luôn đúng)

+) \(x^3+1\ne0\Leftrightarrow x^3\ne-1\Leftrightarrow x\ne-1\)

+) \(x+1\ne0\Leftrightarrow x\ne-1\)

+) \(3x^2+6x\ne0\Leftrightarrow3x\left(x+2\right)\ne0\)

\(\Leftrightarrow x\ne0;x\ne-2\)

+) \(x^2-4\ne0\Leftrightarrow x^2\ne4\Leftrightarrow x\ne\pm2\)

Vậy ĐKXĐ của A là \(x\ne-1;x\ne0;x\ne\pm2\)

a, \(Đkxđ:\hept{\begin{cases}x\ne-1\\x\ne0\\x\ne-2\end{cases}}\)

\(A=\left[\frac{\left(x+1\right)^2}{\left(x+1\right)^2-3x}-\frac{2x^2+4x-1}{x^3+1}-\frac{1}{x+1}\right]:\frac{x^2-4}{3x^2+6x}\)

\(=\left[\frac{x^2+2x+1}{x^2-x+1}-\frac{2x^2+4x-1}{\left(x+1\right)\left(x^2-x+1\right)}-\frac{1}{x+1}\right].\frac{3x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

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

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

\(=\frac{3x}{x-2}=3+\frac{6}{x-2}\)

b, Để A nguyên thì \(\Leftrightarrow6\)chia hết cho \(x-2\)

Hay \(\left(x-2\right)\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

Vậy ............................

b) \(A=\left(\frac{\left(x+1\right)^2}{\left(x+1\right)^2-3x}-\frac{2x^2+4x-1}{x^3+1}-\frac{1}{x+1}\right):\frac{x^2-4}{3x^2+6x}\)

\(=\left(\frac{\left(x+1\right)^2}{x^2-x+1}-\frac{2x^2+4x-1}{x^3+1}-\frac{x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}\right)\)\(:\frac{x^2-4}{3x^2+6x}\)

\(=\left(\frac{\left(x+1\right)^3}{x^3+1}-\frac{2x^2+4x-1}{x^3+1}-\frac{x^2-x+1}{x^3+1}\right)\)\(.\frac{3x^2+6x}{x^2-4}\)

\(=\left(\frac{x^3+3x^2+3x+1}{x^3+1}-\frac{2x^2+4x-1}{x^3+1}-\frac{x^2-x+1}{x^3+1}\right)\)\(.\frac{3x^2+6x}{x^2-4}\)

\(=\frac{x^3+1}{x^3+1}\)\(.\frac{3x^2+6x}{x^2-4}\)\(=\frac{3x^2+6x}{x^2-4}\)

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

A nguyên\(\Leftrightarrow3x⋮x-2\)

\(\Leftrightarrow3\left(x-2\right)+6⋮x-2\)

Mà \(\left(x-2\right)⋮x-2\Rightarrow6⋮x-2\)

\(\Rightarrow x-2\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

Lập bảng:

Vậy\(x\in\left\{3;1;4;0;5;-1;8;-4\right\}\)

Khó vl , dẹp mẹ điiii

a)     \(A=\left(\frac{1}{4}x-y\right)\left(x^2+4xy+16y^2\right)+4\left(4y^3-\frac{1}{16}x^3+1\right)\)

\(\Leftrightarrow A=\frac{1}{4}\left(x-4y\right)\left(x^2+4xy+16y^2\right)+16y^3-\frac{1}{4}x^3+4\)

\(\Leftrightarrow A=\frac{1}{4}\left(x^3-64y^3\right)+16y^3-\frac{1}{4}x^3+4\)

\(\Leftrightarrow A=\frac{1}{4}x^3-16y^3+16y^3-\frac{1}{4}x^3+4\)

\(\Leftrightarrow A=4\)

b) \(B=2x\left(x-4\right)^2-\left(x+5\right)\left(x-2\right)\left(x+2\right)+2\left(x-5\right)^2-\left(x-1\right)^2\)

\(\Leftrightarrow B=2x\left(x^2-8x+16\right)-\left(x+5\right)\left(x^2-4\right)+2\left(x^2-10x+25\right)-\left(x^2-2x+1\right)\)

\(\Leftrightarrow B=2x^3-16x^2+32x-x^3-5x^2+4x+20+2x^2-20x+50-x^2+2x-1\)

\(\Leftrightarrow B=x^3-20x^2+18x+69\)

c) \(C=\frac{80x^3-125x}{3\left(x-3\right)-\left(x-3\right)\left(8-4x\right)}\)

\(\Leftrightarrow C=\frac{5x\left(16x^2-25\right)}{\left(x-3\right)\left(3-8+4x\right)}\)

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

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

\(\Leftrightarrow C=\frac{20x^2+25x}{x-3}\)

d) \(D=\frac{\left(a-b\right)\left(c-d\right)}{\left(b^2-a^2\right)\left(d^2-c^2\right)}\)

\(\Leftrightarrow D=\frac{\left(a-b\right)\left(c-d\right)}{\left(a^2-b^2\right)\left(c^2-d^2\right)}\)

\(\Leftrightarrow D=\frac{\left(a-b\right)\left(c-d\right)}{\left(a-b\right)\left(a+b\right)\left(c-d\right)\left(c+d\right)}\)

\(\Leftrightarrow D=\frac{1}{\left(a+b\right)\left(c+d\right)}\)

Chúc bạn học tốt !

b: \(A=\dfrac{2-1}{3\cdot2}=\dfrac{1}{6}\)

\(ĐKXĐ:x\ne\pm1\)

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

\(\Leftrightarrow A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}-\frac{4x^2}{x^2-1}\right):\frac{2\left(x^2-1\right)}{\left(x-1\right)^2}\)

\(\Leftrightarrow A=\frac{\left(x+1\right)^2-\left(x-1\right)^2-4x^2}{x^2-1}.\frac{\left(x-1\right)^2}{2\left(x^2-1\right)}\)

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

\(\Leftrightarrow A=\frac{4x-4x^2}{x^2-1}.\frac{\left(x-1\right)^2}{2\left(x^2-1\right)}\)

\(\Leftrightarrow A=\frac{-4x\left(x-1\right)^3}{2\left(x-1\right)^2\left(x+1\right)^2}\)

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

b) Thay x = -3 vào A, ta được :

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

\(\Leftrightarrow A=\frac{6.\left(-4\right)}{2^2}\)

\(\Leftrightarrow A=-6\)

c) Để A > -1

\(\Leftrightarrow-2x\left(x-1\right)>-\left(x+1\right)^2\)

\(\Leftrightarrow2x\left(x-1\right)< \left(x+1\right)^2\)

\(\Leftrightarrow2x^2-2x< x^2+2x+1\)

\(\Leftrightarrow x^2-4x-1< 0\)

\(\Leftrightarrow\left(x-2\right)^2-5< 0\)

\(\Leftrightarrow\left(x-2\right)^2< 5\)

Đoạn này bạn tự tìm giá trị x thỏa mãn là xong (Chú ý ĐKXĐ)

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

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

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

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

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

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

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

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

a) ĐKXĐ: \(x\ne-2;x\ne2\), rút gọn:

\(A=\left[\frac{3\left(x-2\right)-2x\left(x+2\right)+2\left(2x^2+3\right)}{2\left(x-2\right)\left(x+2\right)}\right]\div\frac{2x-1}{4\left(x-2\right)}\)

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

b) Ta có: \(\left|x-1\right|=3\Leftrightarrow\hept{\begin{cases}x-1=3\\x-1=-3\end{cases}\Leftrightarrow\hept{\begin{cases}x=4\left(n\right)\\x=-2\left(l\right)\end{cases}}}\)

=> Khi \(x=4\)thì \(A=\frac{4}{4+2}=\frac{4}{6}=\frac{2}{3}\)

c) \(A< 2\Leftrightarrow\frac{4}{x+2}< 2\Leftrightarrow4< 2x+4\Leftrightarrow0< 2x\Leftrightarrow x>0\)Vậy \(A< 2,\forall x>0\)

d) \(\left|A\right|=1\Leftrightarrow\left|\frac{4}{x+2}\right|=1\Leftrightarrow\hept{\begin{cases}\frac{4}{x+2}=1\\\frac{4}{x+2}=-1\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\left(l\right)\\x=-6\left(n\right)\end{cases}}}\)Vậy \(\left|A\right|=1\)khi và chỉ khi x = -6