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a) \(ĐKXĐ:\hept{\begin{cases}x\ne2\\x\ne1\end{cases}}\)
\(A=\frac{2x+1}{x^2-3x+2}+\frac{x+1}{1-x}-\frac{x^2+5}{x^2-3x+2}+\frac{x^2+x}{x-1}\)
\(\Leftrightarrow A=\frac{2x+1}{\left(x-1\right)\left(x-2\right)}-\frac{x+1}{x-1}-\frac{x^2+5}{\left(x-2\right)\left(x-1\right)}+\frac{x^2+x}{x-1}\)
\(\Leftrightarrow A=\frac{2x+1-\left(x+1\right)\left(x-2\right)-x^2-5+\left(x^2+x\right)\left(x-2\right)}{\left(x-1\right)\left(x-2\right)}\)
\(\Leftrightarrow A=\frac{2x+1-x^2+x+2-x^2-5+x^3-x^2-2x}{\left(x-1\right)\left(x-2\right)}\)
\(\Leftrightarrow A=\frac{x^3-3x^2+x-2}{\left(x-1\right)\left(x-2\right)}\)
b) Khi \(x^2-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)=.0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\left(ktm\right)\\x=-1\left(tm\right)\end{cases}}\)
\(\Leftrightarrow A=\frac{\left(-1\right)^3-3\left(-1\right)^2-1-2}{\left(-1-2\right)\left(-1-1\right)}=\frac{\left(-1\right)-3-1-2}{\left(-3\right)\left(-2\right)}=\frac{7}{6}\)
c) Để A = 0
\(\Leftrightarrow\frac{x^3-3x^2+x-2}{\left(x-1\right)\left(x-2\right)}=0\)
\(\Leftrightarrow x^3-3x^2+x-2=0\)2.89328919
Phần này mik k biết phân tích như thế nào, tính ra :
\(\Leftrightarrow x\approx2,89328919\)
Nhưng nếu đề bắt tìm nghiệm nguyên của x thì \(S=\varnothing\)nhé !
d) Để \(A\inℤ\)
\(\Leftrightarrow x^3-3x^2+x-2⋮\left(x-2\right)\left(x-1\right)\)
\(\Leftrightarrow\hept{\begin{cases}x^3-3x^2+x-2⋮x-2\\x^3-3x+x-2⋮x-1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\left(x^2-x-1\right)\left(x-2\right)-4⋮x-2\\\left(x^2-2x-1\right)\left(x-1\right)-3⋮x-1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}4⋮x-2\\3⋮x-1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x-2\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\\x-1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\in\left\{1;3;0;4;-2;6\right\}\\x\in\left\{0;2;-2;4\right\}\end{cases}}\)
\(\Leftrightarrow x\in\left\{0;-2;4\right\}\)
Vậy để \(A\inℤ\Leftrightarrow x\in\left\{0;-2;4\right\}\)
\(\left(x+4\right)^2-81=0\Leftrightarrow\left(x+4\right)^2-9^2=0\)
\(\Leftrightarrow\left(x+4+9\right)\times\left(x+4-9\right)=0\)
\(\Leftrightarrow\left(x+13\right)\times\left(x-5\right)=0\)
\(\left[{}\begin{matrix}x+13=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-13\\x=5\end{matrix}\right.\)
a: \(A=\dfrac{2x+1}{\left(x-1\right)\left(x-2\right)}+\dfrac{x+1}{x-1}-\dfrac{x^2+5}{\left(x-1\right)\left(x-2\right)}+\dfrac{x^2+x}{x-1}\)
\(=\dfrac{2x+1-x^2-5}{\left(x-1\right)\left(x-2\right)}+\dfrac{x+1+x^2+x}{x-1}\)
\(=\dfrac{-x^2+2x-4+\left(x^2+2x+1\right)\left(x-2\right)}{\left(x-1\right)\left(x-2\right)}\)
\(=\dfrac{-x^2+2x-4+x^3-2x^2+2x^2-4x+x-2}{\left(x-1\right)\left(x-2\right)}\)
\(=\dfrac{x^3-x^2-x-6}{\left(x-1\right)\left(x-2\right)}\)
b: Để A là số nguyên thì \(x^3-3x^2+2x+2x^2-6x+4+3x-10⋮\left(x-1\right)\left(x-2\right)\)
=>\(3x-10⋮x^2-3x+2\)
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