\(B=\left[\dfrac{3}{x+1}+\left(\dfrac{3}{x+1}-\dfrac{x}{x^2+2x+1}\right):\dfrac{2x^2+3x}{x+1}\right]:\dfrac{1+3x}{x^2+x}\)
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chứng minh rằng :
a) left(dfrac{x^2-2x}{2x^2+8}-dfrac{2x^2}{8-4x+2x^2-x}right)left(1-dfrac{1}{x}-dfrac{2}{x^2}right)dfrac{x+1}{2x}
b)left[dfrac{2}{3x}-dfrac{2}{x+1}left(dfrac{x+1}{3x}-x-1right)right]:dfrac{x+1}{x}dfrac{2x}{x-1}
c)left[dfrac{2}{left(x+1right)^3}left(dfrac{1}{x}+1right)+dfrac{1}{x^2+2x+1}left(dfrac{1}{x^2}+1right)right]:dfrac{x-1}{x^3}dfrac{x}{x-1}
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chứng minh rằng :
a) \(\left(\dfrac{x^2-2x}{2x^2+8}-\dfrac{2x^2}{8-4x+2x^2-x}\right)\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)=\dfrac{x+1}{2x}\)
b)\(\left[\dfrac{2}{3x}-\dfrac{2}{x+1}\left(\dfrac{x+1}{3x}-x-1\right)\right]:\dfrac{x+1}{x}=\dfrac{2x}{x-1}\)
c)\(\left[\dfrac{2}{\left(x+1\right)^3}\left(\dfrac{1}{x}+1\right)+\dfrac{1}{x^2+2x+1}\left(\dfrac{1}{x^2}+1\right)\right]:\dfrac{x-1}{x^3}=\dfrac{x}{x-1}\)
b: \(=\left[\dfrac{2}{3x}-\dfrac{2}{x+1}\cdot\dfrac{x+1-3x^2-3x}{3x}\right]\cdot\dfrac{x}{x+1}\)
\(=\left(\dfrac{2}{3x}-\dfrac{2}{x+1}\cdot\dfrac{-3x^2-2x+1}{3x}\right)\cdot\dfrac{x}{x+1}\)
\(=\dfrac{2x+2+6x^2+4x-2}{3x\left(x+1\right)}\cdot\dfrac{x}{x+1}\)
\(=\dfrac{6x^2+6x}{3\left(x+1\right)}\cdot\dfrac{1}{x+1}\)
\(=\dfrac{6x\left(x+1\right)}{3\left(x+1\right)^2}=\dfrac{2x}{x+1}\)
c: \(VT=\left[\dfrac{2}{\left(x+1\right)^3}\cdot\dfrac{x+1}{x}+\dfrac{1}{\left(x+1\right)^2}\cdot\dfrac{1+x^2}{x^2}\right]\cdot\dfrac{x^3}{x-1}\)
\(=\left(\dfrac{2}{x\left(x+1\right)^2}+\dfrac{x^2+1}{x^2\cdot\left(x+1\right)^2}\right)\cdot\dfrac{x^3}{x-1}\)
\(=\dfrac{2x+x^2+1}{x^2\cdot\left(x+1\right)^2}\cdot\dfrac{x^3}{x-1}\)
\(=\dfrac{\left(x+1\right)^2}{\left(x+1\right)^2}\cdot\dfrac{x}{x-1}=\dfrac{x}{x-1}\)
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giải phương trình
a.\(\left(2x-3\right)^2=\left(2x-3\right)\left(x+1\right)\)
b.\(x\left(2x-9\right)=3x\left(x-5\right)\)
c.\(3x-15=2x\left(x-5\right)\)
d.\(\dfrac{5-x}{2}=\dfrac{3x-4}{6}\)
e.\(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)
a) Ta có: \(\left(2x-3\right)^2=\left(2x-3\right)\left(x+1\right)\)
\(\Leftrightarrow\left(2x-3\right)^2-\left(2x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(2x-3-x-1\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=4\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{3}{2};4\right\}\)
b) Ta có: \(x\left(2x-9\right)=3x\left(x-5\right)\)
\(\Leftrightarrow x\left(2x-9\right)-3x\left(x-5\right)=0\)
\(\Leftrightarrow x\left(2x-9\right)-x\left(3x-15\right)=0\)
\(\Leftrightarrow x\left(2x-9-3x+15\right)=0\)
\(\Leftrightarrow x\left(6-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
Vậy: S={0;6}
c) Ta có: \(3x-15=2x\left(x-5\right)\)
\(\Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\3-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\2x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{5;\dfrac{3}{2}\right\}\)
d) Ta có: \(\dfrac{5-x}{2}=\dfrac{3x-4}{6}\)
\(\Leftrightarrow6\left(5-x\right)=2\left(3x-4\right)\)
\(\Leftrightarrow30-6x=6x-8\)
\(\Leftrightarrow30-6x-6x+8=0\)
\(\Leftrightarrow-12x+38=0\)
\(\Leftrightarrow-12x=-38\)
\(\Leftrightarrow x=\dfrac{19}{6}\)
Vậy: \(S=\left\{\dfrac{19}{6}\right\}\)
e) Ta có: \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{12x}{6}+\dfrac{10}{6}\)
\(\Leftrightarrow6x+4-3x-1=12x+10\)
\(\Leftrightarrow3x+3-12x-10=0\)
\(\Leftrightarrow-9x-7=0\)
\(\Leftrightarrow-9x=7\)
\(\Leftrightarrow x=-\dfrac{7}{9}\)
Vậy: \(S=\left\{-\dfrac{7}{9}\right\}\)
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Thực hiện phép tínha,left(dfrac{1}{x^2+x}-dfrac{2-x}{x+1}right):left(dfrac{1}{x}+x-2right)b,left(dfrac{3x}{1-3x}+dfrac{2x}{3x+1}right):dfrac{6x^2+10x}{1-6x+9x^2}c,left(dfrac{9}{x^3-9x}+dfrac{1}{x+3}right):left(dfrac{x-3}{x^2+3x}-dfrac{x}{3x+9}right)d,dfrac{x+1}{x+2}:left(dfrac{x+2}{x+3}:dfrac{x+3}{x+1}right)e,dfrac{8}{left(x^2+3right)left(x^2-1right)}+dfrac{2}{x^2+3}+dfrac{1}{x+1}f,dfrac{x+y}{2left(x-yright)}-dfrac{x-y}{2left(x+yright)}+dfrac{2y^2}{x^2-y^2}g,dfrac{x-1}{x^3}-dfrac{x+1}{x^3-x^2}+d...
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Thực hiện phép tính
\(a,\left(\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)
\(b,\left(\dfrac{3x}{1-3x}+\dfrac{2x}{3x+1}\right):\dfrac{6x^2+10x}{1-6x+9x^2}\)
\(c,\left(\dfrac{9}{x^3-9x}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\)
\(d,\dfrac{x+1}{x+2}:\left(\dfrac{x+2}{x+3}:\dfrac{x+3}{x+1}\right)\)
\(e,\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2}{x^2+3}+\dfrac{1}{x+1}\)
\(f,\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{x^2-y^2}\)
\(g,\dfrac{x-1}{x^3}-\dfrac{x+1}{x^3-x^2}+\dfrac{3}{x^3-2x^2+x}\)
\(h,\dfrac{x^3}{x-1}-\dfrac{x^2}{x+1}-\dfrac{1}{x-1}+\dfrac{1}{x+1}\)
a) Ta có: \(\left(\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)
\(=\left(\dfrac{1}{x\left(x+1\right)}+\dfrac{x+2}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)
\(=\dfrac{x^2+2x+1}{x\left(x+1\right)}:\dfrac{x^2-2x+1}{x}\)
\(=\dfrac{\left(x+1\right)^2}{x\left(x+1\right)}\cdot\dfrac{x}{\left(x-1\right)^2}\)
\(=\dfrac{x+1}{\left(x-1\right)^2}\)
b) Ta có: \(\left(\dfrac{3x}{1-3x}+\dfrac{2x}{3x+1}\right):\dfrac{6x^2+10x}{1-6x+9x^2}\)
\(=\dfrac{3x\left(3x+1\right)+2x\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}:\dfrac{2x\left(3x+5\right)}{\left(1-3x\right)^2}\)
\(=\dfrac{9x^2+3x+2x-6x^2}{\left(1-3x\right)\left(1+3x\right)}:\dfrac{2x\left(3x+5\right)}{\left(1-3x\right)^2}\)
\(=\dfrac{3x^2+5x}{\left(1-3x\right)\left(1+3x\right)}\cdot\dfrac{\left(1-3x\right)^2}{2x\left(3x+5\right)}\)
\(=\dfrac{x\left(3x+5\right)}{1+3x}\cdot\dfrac{1-3x}{2x\left(3x+5\right)}\)
\(=\dfrac{2\left(1-3x\right)}{3x+1}\)
c) Ta có: \(\left(\dfrac{9}{x^3-9x}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\)
\(=\left(\dfrac{9}{x\left(x-3\right)\left(x+3\right)}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x\left(x+3\right)}-\dfrac{x}{3\left(x+3\right)}\right)\)
\(=\dfrac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}:\dfrac{3\left(x-3\right)-x^2}{3x\left(x+3\right)}\)
\(=\dfrac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}\cdot\dfrac{3x\left(x+3\right)}{3x-9-x^2}\)
\(=\dfrac{x^2-3x+9}{x-3}\cdot\dfrac{3}{-\left(x^2-3x+9\right)}\)
\(=\dfrac{-3}{x-3}\)
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1. dfrac{5left(x-1right)+2}{6}-dfrac{7x-1}{4}dfrac{2left(2x+1right)}{7}-5
2. x-dfrac{3left(x+30right)}{15}-24dfrac{1}{2}dfrac{7x}{10}-dfrac{2left(10x+2right)}{5}
3. 14dfrac{1}{2}-dfrac{2left(x+3right)}{5}dfrac{3x}{2}-dfrac{2left(x-7right)}{3}
4. dfrac{x+1}{3}+dfrac{3left(2x+1right)}{4}dfrac{2x+3left(x+1right)}{6}+dfrac{7+12x}{12}
5. dfrac{3left(2x-1right)}{4}-dfrac{3x+1}{10}+1dfrac{2left(3x+2right)}{5}
6. x-dfrac{3}{17}left(2x-1right)dfrac{7}{34}left(1-2xright)+dfrac{10x-3}{2}
7. dfrac{3le...
Đọc tiếp
1. \(\dfrac{5\left(x-1\right)+2}{6}-\dfrac{7x-1}{4}=\dfrac{2\left(2x+1\right)}{7}-5\)
2. \(x-\dfrac{3\left(x+30\right)}{15}-24\dfrac{1}{2}=\dfrac{7x}{10}-\dfrac{2\left(10x+2\right)}{5}\)
3. \(14\dfrac{1}{2}-\dfrac{2\left(x+3\right)}{5}=\dfrac{3x}{2}-\dfrac{2\left(x-7\right)}{3}\)
4. \(\dfrac{x+1}{3}+\dfrac{3\left(2x+1\right)}{4}=\dfrac{2x+3\left(x+1\right)}{6}+\dfrac{7+12x}{12}\)
5. \(\dfrac{3\left(2x-1\right)}{4}-\dfrac{3x+1}{10}+1=\dfrac{2\left(3x+2\right)}{5}\)
6. \(x-\dfrac{3}{17}\left(2x-1\right)=\dfrac{7}{34}\left(1-2x\right)+\dfrac{10x-3}{2}\)
7. \(\dfrac{3\left(x-3\right)}{4}+\dfrac{4x-10,5}{10}=\dfrac{3\left(x+1\right)}{5}+6\)
8. \(\dfrac{2\left(3x+1\right)+1}{4}-5=\dfrac{2\left(3x-1\right)}{5}-\dfrac{3x+2}{10}\)
Rút gọn các biểu thức sau : A 2x^2left(-3x^3+2x^2+x-1right)+2xleft(x^2-3x+1right)B 2x:dfrac{1}{2}x+x^2C left[1:left(1+xright)+2x:left(1-x^2right)right]:left(dfrac{1}{x}-1right)D dfrac{x^2-y^2}{x+y}.dfrac{left(x+yright)^2}{x}+dfrac{y^2}{x+y}.dfrac{left(x+yright)^2}{x}E dfrac{left|x-3right|}{x^2-9}.left(x^2+6x+9right)F dfrac{sqrt{x}}{sqrt{x}-5}-dfrac{10sqrt{x}}{x-25}-dfrac{5}{sqrt{x}+5}
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Rút gọn các biểu thức sau :
A = \(2x^2\left(-3x^3+2x^2+x-1\right)+2x\left(x^2-3x+1\right)\)
B = \(2x:\dfrac{1}{2}x+x^2\)
C = \(\left[1:\left(1+x\right)+2x:\left(1-x^2\right)\right]:\left(\dfrac{1}{x}-1\right)\)
D = \(\dfrac{x^2-y^2}{x+y}.\dfrac{\left(x+y\right)^2}{x}+\dfrac{y^2}{x+y}.\dfrac{\left(x+y\right)^2}{x}\)
E = \(\dfrac{\left|x-3\right|}{x^2-9}.\left(x^2+6x+9\right)\)
F = \(\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{10\sqrt{x}}{x-25}-\dfrac{5}{\sqrt{x}+5}\)
Rút gọn các biểu thức sau : A 2x^2left(-3x^3+2x^2+x-1right)+2xleft(x^2-3x+1right)B 2x:dfrac{1}{2}x+x^2C left[1:left(1+xright)+2x:left(1-x^2right)right]:left(dfrac{1}{x}-1right)D dfrac{x^2-y^2}{x+y}.dfrac{left(x+yright)^2}{x}+dfrac{y^2}{x+y}.dfrac{left(x+yright)^2}{x}E dfrac{left|x-3right|}{x^2-9}.left(x^2+6x+9right)F dfrac{sqrt{x}}{sqrt{x}-5}-dfrac{10sqrt{x}}{x-25}-dfrac{5}{sqrt{x}+5}
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Rút gọn các biểu thức sau :
A = \(2x^2\left(-3x^3+2x^2+x-1\right)+2x\left(x^2-3x+1\right)\)
B = \(2x:\dfrac{1}{2}x+x^2\)
C = \(\left[1:\left(1+x\right)+2x:\left(1-x^2\right)\right]:\left(\dfrac{1}{x}-1\right)\)
D = \(\dfrac{x^2-y^2}{x+y}.\dfrac{\left(x+y\right)^2}{x}+\dfrac{y^2}{x+y}.\dfrac{\left(x+y\right)^2}{x}\)
E = \(\dfrac{\left|x-3\right|}{x^2-9}.\left(x^2+6x+9\right)\)
F = \(\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{10\sqrt{x}}{x-25}-\dfrac{5}{\sqrt{x}+5}\)
giải phương trình
a, \(\dfrac{3}{2x-1}+1=\dfrac{2x-1}{2x+1}\)
b,\(\dfrac{3x-1}{x-1}-\dfrac{2x+5}{x+3}+\dfrac{4}{x^2+2x-3}=1\)
c,\(\dfrac{5}{x^2+x-6}-\dfrac{2}{x^2+4x+3}=\dfrac{-3}{2x-1}\)
d, \(\left(x^2-4\right)\left(2x+3\right)=\left(x^2-4\right)\left(x-1\right)\)
e, \(x^3+x^2+x+1=0\)
\(a,\dfrac{3}{2x-1}+1=\dfrac{2x-1}{2x+1};ĐKXĐ:x\ne\pm\dfrac{1}{2}\\ \Leftrightarrow\dfrac{3}{2x-1}-\dfrac{2x-1}{2x+1}+1=0\\ \Leftrightarrow\dfrac{3\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}-\dfrac{\left(2x-1\right)\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}+\dfrac{\left(2x-1\right)\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}=0\\ \Rightarrow3\left(2x+1\right)-\left(2x-1\right)^2+\left(2x-1\right)\left(2x+1\right)=0\\ \Leftrightarrow6x+3-\left(4x^2-4x+1\right)+\left(4x^2-1\right)=0\\ \Leftrightarrow6x+3-4x^2+4x-1+4x^2-1=0\\ \Leftrightarrow10x+1=0\\ \Leftrightarrow10x=-1\\ \Leftrightarrow x=-\dfrac{1}{10}\)
Vậy \(x\in\left\{-\dfrac{1}{10}\right\}\)
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1) dfrac{7x-3}{x-1} dfrac{2}{3}2) dfrac{2left(3-7xright)}{1+x} dfrac{1}{2}3) dfrac{x^{2^{ }}-6}{x} x + dfrac{3}{2}4) dfrac{5}{3x+2} 2x - 15) dfrac{left(x^2+2xright)-left(3x+6right)}{x-3} 06) dfrac{1}{x-2} + 3 dfrac{3-x}{x-2}
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1) \(\dfrac{7x-3}{x-1}\) = \(\dfrac{2}{3}\)
2) \(\dfrac{2\left(3-7x\right)}{1+x}\) = \(\dfrac{1}{2}\)
3) \(\dfrac{x^{2^{ }}-6}{x}\) = x + \(\dfrac{3}{2}\)
4) \(\dfrac{5}{3x+2}\) = 2x - 1
5) \(\dfrac{\left(x^2+2x\right)-\left(3x+6\right)}{x-3}\) = 0
6) \(\dfrac{1}{x-2}\) + 3 = \(\dfrac{3-x}{x-2}\)
1/ ĐKXĐ : \(x\ne1\)
\(\dfrac{7x-3}{x-1}=\dfrac{2}{3}\)
\(\Leftrightarrow21x-9=2x-2\)
\(\Leftrightarrow19x=7\Leftrightarrow x=\dfrac{7}{19}\left(tm\right)\)
Vậy...
b/ \(\dfrac{2\left(3-7x\right)}{1+x}=\dfrac{1}{2}\) ĐKXĐ : \(x\ne-1\)
\(\Leftrightarrow12-28x=1+x\)
\(\Leftrightarrow11=29x\Leftrightarrow x=\dfrac{11}{29}\) \(\left(tm\right)\)
Vậy....
c/ ĐKXĐ : \(x\ne0\)
\(\dfrac{x^2-6}{x}=x+\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{x^2-6}{x}=\dfrac{2x+3}{2}\)
\(\Leftrightarrow2x^2-12=2x^2+3x\)
\(\Leftrightarrow3x=-12\Leftrightarrow x=-4\) \(\left(tm\right)\)
Vậy...
4/ ĐKXĐ : \(x\ne-\dfrac{2}{3}\)
\(\dfrac{5}{3x+2}=2x-1\)
\(\Leftrightarrow\left(2x-1\right)\left(3x+2\right)=5\)
\(\Leftrightarrow6x^2+4x-3x-2=5\)
\(\Leftrightarrow6x^2+x-7=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{6}\\x=1\end{matrix}\right.\)
Vậy....
5,6 Tương tự nhé !
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1)ĐKXĐ: \(x\ne1\)
Ta có: \(\dfrac{7x-3}{x-1}=\dfrac{2}{3}\)
\(\Leftrightarrow3\left(7x-3\right)=2\left(x-1\right)\)
\(\Leftrightarrow21x-9=2x-2\)
\(\Leftrightarrow21x-9-2x+2=0\)
\(\Leftrightarrow19x-7=0\)
\(\Leftrightarrow19x=7\)
\(\Leftrightarrow x=\dfrac{7}{19}\)(nhận)
Vậy: \(S=\left\{\dfrac{7}{19}\right\}\)
2) ĐKXĐ: \(x\ne-1\)
Ta có: \(\dfrac{2\left(3-7x\right)}{1+x}=\dfrac{1}{2}\)
\(\Leftrightarrow4\left(3-7x\right)=x+1\)
\(\Leftrightarrow12-28x-x-1=0\)
\(\Leftrightarrow-29x+11=0\)
\(\Leftrightarrow-29x=-11\)
\(\Leftrightarrow x=\dfrac{11}{29}\)
Vậy: \(S=\left\{\dfrac{11}{29}\right\}\)
3) ĐKXĐ: \(x\ne0\)
Ta có: \(\dfrac{x^2-6}{x}=x+\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{x^2-6}{x}=\dfrac{2x+3}{2}\)
\(\Leftrightarrow2\left(x^2-6\right)=x\left(2x+3\right)\)
\(\Leftrightarrow2x^2-12=2x^2+6x\)
\(\Leftrightarrow2x^2-12-2x^2-6x=0\)
\(\Leftrightarrow-6x-12=0\)
\(\Leftrightarrow-6x=12\)
\(\Leftrightarrow x=-2\)
Vậy: S={-2}
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BÀI 1 : RÚT GỌN CÁC BIỂU THỨC SAU .
a, \(\dfrac{3}{x-3}-\dfrac{6x}{9-x^2}+\dfrac{x}{x+3}\)
b, \(\left(\dfrac{3x}{1-3x}+\dfrac{2x}{3x+1}\right):\dfrac{6x^2+10x}{9x^2-6x+1}\)
c, \(\left(\dfrac{9}{x^3-9x}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\)
d, \(\dfrac{1-x^2}{x}\left(\dfrac{x^2}{x+3}-1\right)+\dfrac{3x^2-14x+3}{x^2+3x}\)
câu d
\(D=\dfrac{\left(1-x^2\right)}{x}\left(\dfrac{x^2}{x+3}-1\right)+\dfrac{3x^2-14x+3}{x^2+3x}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{\left(1-x^2\right)\left(x^2-x-3\right)+3x^2-14x+3}{x\left(x+3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{x^2-x-3-x^4+x^3-3x^2+3x^2-14x+3}{x\left(x+3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-x^4+x^3+x^2-15x}{x\left(x+3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-x\left(x^3-x^2-x+15\right)}{x\left(x+3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-\left(x^3-x^2-x+15\right)}{\left(x+3\right)}\end{matrix}\right.\)
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