a) Do \(\left|x\right|\ge0\)

\(\Rightarrow A=\left|x\right|+5\ge5\)

\(minA=5\Leftrightarrow x=0\)

b) Do \(\left|x-\dfrac{2}{3}\right|\ge0\)

\(\Rightarrow B=\left|x-\dfrac{2}{3}\right|-4\ge-4\)

\(minB=-4\Leftrightarrow x=\dfrac{2}{3}\)

c) Do \(\left|3x-1\right|\ge0\)

\(\Rightarrow C=\left|3x-1\right|-\dfrac{1}{2}\ge-\dfrac{1}{2}\)

\(minC=-\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{3}\)

\(A=\left|x\right|+5\ge5\)

Dấu \("="\Leftrightarrow x=0\)

\(B=\left|x-\dfrac{2}{3}\right|-4\ge-4\)

Dấu \("="\Leftrightarrow x-\dfrac{2}{3}=0\Leftrightarrow x=\dfrac{2}{3}\)

\(C=\left|3x-1\right|-\dfrac{1}{2}\ge-\dfrac{1}{2}\)

Dấu \("="\Leftrightarrow3x-1=0\Leftrightarrow x=\dfrac{1}{3}\)

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

a)Đa thức B có nghĩa\(\Leftrightarrow x+1\ne0\)và\(x-1\ne0\)và\(x\ne0\Leftrightarrow x\ne-1\)và\(x\ne1\)và\(x\ne0\)

b)Ta có:\(B=\left(\frac{x^2+1}{x+1}-1\right)\left(\frac{4}{x-1}-\frac{2}{x}\right)=\left(\frac{x^2+1}{x+1}-\frac{x+1}{x+1}\right)\left(\frac{4.x}{\left(x-1\right).x}-\frac{2.\left(x-1\right)}{x.\left(x-1\right)}\right)\)

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

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

a) B có nghĩa \(\Leftrightarrow\hept{\begin{cases}x+1\ne0\\x-1\ne0\\x\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne-1\\x\ne1\\x\ne0\end{cases}}\)

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

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

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

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

\(a,\frac{x+22}{x+1}\inℤ\Leftrightarrow x+22⋮x+1\)

\(\Rightarrow x+1+21⋮x+1\) 

     \(x+1⋮x+1\)

\(\Rightarrow21⋮x+1\)

\(\Rightarrow x+1\inƯ\left(21\right)\)

\(\Rightarrow x+1\in\left\{-1;1;-3;3;-7;7;-21;21\right\}\)

\(\Rightarrow x\in\left\{-2;0;-4;2;-8;6;-22;20\right\}\)

vậy___ 

\(b,\frac{3x+1}{2x+1}\inℤ\Leftrightarrow3x+1⋮2x+1\)

\(\Rightarrow2\left(3x+1\right)⋮2x+1\)

\(\Rightarrow6x+2⋮2x+1\)

\(\Rightarrow6x+2+1-1⋮2x+1\)

\(\Rightarrow6x+3-1⋮2x+1\)

\(\Rightarrow3\left(2x+1\right)-1⋮2x+1\)

      \(3\left(2x+1\right)⋮2x+1\)

\(\Rightarrow1⋮2x+1\)

\(\Rightarrow2x+1\inƯ\left(1\right)\)

đến đây lm như phần a

\(c,\frac{2x+1}{6-n}\inℤ\Leftrightarrow2x+1⋮6-n\)

\(\Rightarrow2x+1+11-11⋮6-n\)

\(\Rightarrow2x+12-11⋮6-n\)

\(\Rightarrow2\left(x+6\right)-11⋮6-n\)

      \(2\left(x+6\right)⋮6-n\)

\(\Rightarrow11⋮6-n\)

tự lm tp

phần c thì k chắc lắm

cảm ơn nhé

B=|3-x|+|x+4|>=|3-x+x+4|=7

Dấu = xảy ra khi -4<=x<=3