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

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

D xác định\(\Leftrightarrow\left(x^2+1\right)\left(x+1\right)\ne0\)

Mà \(x^2+1>0\)nên \(x+1\ne0\Leftrightarrow x\ne-1\)

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

c) D nguyên \(\Leftrightarrow\frac{3}{x^2+1}\in Z\Leftrightarrow3⋮\left(x^2+1\right)\)

\(\Leftrightarrow x^2+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

Mà \(x^2+1>0\)nên \(x^2+1\in\left\{1;3\right\}\)

\(TH1:x^2+1=1\Leftrightarrow x^2=0\Leftrightarrow x=0\)

\(TH2:x^2+1=3\Leftrightarrow x^2=2\Leftrightarrow x=\pm\sqrt{2}\)

Vậy D nguyên \(\Leftrightarrow x\in\left\{0;\pm\sqrt{2}\right\}\)

d) Ta có: \(x^2\ge0\)

\(\Leftrightarrow x^2+1\ge1\)

\(\Leftrightarrow\frac{3}{x^2+1}\le3\)

Vậy D_max = 3\(\Leftrightarrow x^2+1=1\Leftrightarrow x=0\)

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

Bổ sung phần c và d luôn:

c, C = \(\dfrac{2}{5}\)

\(\Leftrightarrow\) \(\dfrac{x^2-1}{2x^2+3}\) = \(\dfrac{2}{5}\)

\(\Leftrightarrow\) 5(x² - 1) = 2(2x² + 3)

\(\Leftrightarrow\) 5x² - 5 = 4x² + 6

\(\Leftrightarrow\) x² = 11

\(\Leftrightarrow\) x² - 11 = 0

\(\Leftrightarrow\) (x - \(\sqrt{11}\))(x + \(\sqrt{11}\)) = 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x-\sqrt{11}=0\\x+\sqrt{11}=0\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=\sqrt{11}\left(TM\right)\\x=-\sqrt{11}\left(TM\right)\end{matrix}\right.\)

d, Ta có: \(\dfrac{x^2-1}{2x^2+3}\) = \(\dfrac{x^2+\dfrac{3}{2}-\dfrac{5}{2}}{2\left(x^2+\dfrac{3}{2}\right)}\) = \(\dfrac{1}{2}\) - \(\dfrac{5}{4\left(x^2+\dfrac{3}{2}\right)}\)

C nguyên \(\Leftrightarrow\) \(\dfrac{5}{4\left(x^2+\dfrac{3}{2}\right)}\) nguyên \(\Leftrightarrow\) 5 \(⋮\) 4(x² + \(\dfrac{3}{2}\))

\(\Leftrightarrow\) 4(x² + \(\dfrac{3}{2}\)) \(\in\) Ư(5)

Xét các TH:

4(x² + \(\dfrac{3}{2}\)) = 5 \(\Leftrightarrow\) x² = \(\dfrac{-1}{4}\) \(\Leftrightarrow\) x² + \(\dfrac{1}{4}\) = 0 (Vô nghiệm)

4(x² + \(\dfrac{3}{2}\)) = -5 \(\Leftrightarrow\) x² = \(\dfrac{-11}{4}\) \(\Leftrightarrow\) x² + \(\dfrac{11}{4}\) = 0 (Vô nghiệm)

4(x² + \(\dfrac{3}{2}\)) = 1 \(\Leftrightarrow\) x² = \(\dfrac{-5}{4}\) \(\Leftrightarrow\) x² + \(\dfrac{5}{4}\) = 0 (Vô nghiệm)

4(x² + \(\dfrac{3}{2}\)) = -1 \(\Leftrightarrow\) x² = \(\dfrac{-7}{4}\) \(\Leftrightarrow\) x² + \(\dfrac{7}{4}\) = 0 (Vô nghiệm)

Vậy không có giá trị nào của x \(\in\) Z thỏa mãn C \(\in\) Z

Chúc bn học tốt! (Ko bt đề sai hay ko nữa :v)

a, ĐK : \(x\ne\pm3;\frac{1}{2}\)

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

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

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

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

b, Ta có : \(\left|x+1\right|=\frac{1}{2}\)

TH1 : \(x+1=\frac{1}{2}\Leftrightarrow x=-\frac{1}{2}\)

Thay vào biểu thức A ta được : \(\frac{-1+1}{-\frac{1}{2}+3}=0\)

TH2 : \(x+1=-\frac{1}{2}\Leftrightarrow x=-\frac{3}{2}\)

Thay vào biểu thức A ta được : \(\frac{-3+1}{-\frac{3}{2}+3}=\frac{-2}{\frac{3}{2}}=-\frac{4}{3}\)

c, Ta có : \(P=\frac{x}{2}\Rightarrow\frac{2x+1}{x+3}=\frac{x}{2}\Rightarrow4x+2=x^2+3x\)

\(\Leftrightarrow x^2-x-2=0\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow x=-1;x=2\)

b, Ta có : \(\frac{2x+1}{x+3}=\frac{2\left(x+3\right)-5}{x+3}=2-\frac{5}{x+3}\)

\(\Rightarrow x+3\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

a) \(-ĐKXĐ:x\ne\pm2;1\)

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

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

\(=\left[\frac{x-2}{\left(x-2\right)\left(x+2\right)}+\frac{\left(-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x}{\left(x-2\right)\left(x+2\right)}\right]\)\(.\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)

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

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

b) \(A>0\Leftrightarrow\frac{x+1}{\left(x+2\right)^2}>0\Leftrightarrow\orbr{\begin{cases}x+1< 0;\left(x+2\right)^2< 0\left(voly\right)\\x+1>0;\left(x+2\right)^2>0\end{cases}}\)

\(\Leftrightarrow x>1;x>-2\Leftrightarrow x>1\)

Vậy với mọi x thỏa mãn x>1 thì A > 0

c) Ta có : \(x^2+3x+2=0\Leftrightarrow x^2+x+2x+2=0\)

\(\Leftrightarrow x\left(x+1\right)+2\left(x+1\right)=0\Leftrightarrow\left(x+1\right)\left(x+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-2\end{cases}}\)

Vậy x = -1;-2