a, ĐKXĐ: \(\hept{\begin{cases}x^3+1\ne0\\x^9+x^7-3x^2-3\ne0\\x^2+1\ne0\end{cases}}\)

b, \(Q=\left[\left(x^4-x+\frac{x-3}{x^3+1}\right).\frac{\left(x^3-2x^2+2x-1\right)\left(x+1\right)}{x^9+x^7-3x^2-3}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)

\(Q=\left[\frac{\left(x^3+1\right)\left(x^4-x\right)+x-3}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)

\(Q=\left[\left(x^7-3\right).\frac{\left(x-1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)

\(Q=\frac{x-1+x^2+1-2x-12}{x^2+1}\)

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

a/ Đkxđ: \(\left\{{}\begin{matrix}x\ne0\\x+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne-1\end{matrix}\right.\)

Vậy phân thức được xác định khi \(\left\{{}\begin{matrix}x\ne0\\x\ne-1\end{matrix}\right.\)

b/ \(A=\left[1+\frac{1}{x}+\frac{2}{x+1}\left(1+\frac{1}{x}\right)\right]:\frac{x^3+27}{2x}\)

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

\(=\left[\frac{x\left(x+1\right)+\left(x+1\right)+2x+2}{\left(x+1\right)x}\right].\frac{2x}{\left(x+3\right)\left(x^2-3x+9\right)}\)

\(=\frac{x^2+4x+3}{\left(x+1\right)x}.\frac{2x}{\left(x+3\right)\left(x^2-3x+9\right)}=\frac{\left(x+1\right)\left(x+3\right)}{\left(x+1\right)x}.\frac{2x}{\left(x+3\right)\left(x^2-3x+9\right)}\)

\(=\frac{2}{x^2-3x+9}\)

Câu 1 :

a) ĐKXĐ : \(\hept{\begin{cases}x+1\ne0\\2x-6\ne0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x\ne-1\\x\ne3\end{cases}}\)

b) Để \(P=1\Leftrightarrow\frac{4x^2+4x}{\left(x+1\right)\left(2x-6\right)}=1\)

\(\Leftrightarrow\frac{4x^2+4x-\left(x+1\right)\left(2x-6\right)}{\left(x+1\right)\left(2x-6\right)}=0\)

\(\Rightarrow4x^2+4x-2x^2+4x+6=0\)

\(\Leftrightarrow2x^2+8x+6=0\)

\(\Leftrightarrow x^2+4x+4-1=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+3=0\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=-1\left(KTMĐKXĐ\right)\\x=-3\left(TMĐKXĐ\right)\end{cases}}\)

Vậy : \(x=-3\) thì P = 1.

a) \(ĐKXĐ:x\ne1\)

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

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

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

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

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

\(=\frac{1}{x-1}\)

c) Với \(\forall x\)(\(x\ne1\)) thì biểu thức được xác định .

P/s : Theo mik câu c nên chuyển thành : Tìm x để biểu thức đạt giá trị nguyên.

Tại thấy câu c k khác j câu a !

a) ĐKXĐ: \(x\ne-1;0;1.\)Ta có:

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

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

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

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

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

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

Vậy \(A=\frac{x}{x-1}\)với \(x\ne-1;0;1.\)

b) A < 1 \(\Leftrightarrow\frac{x}{x-1}< 1\Leftrightarrow\frac{x}{x-1}-1< 0\Leftrightarrow\frac{x}{x-1}-\frac{x-1}{x-1}< 0\)\(\Leftrightarrow\frac{1}{x-1}< 0\)

\(\Leftrightarrow x-1< 0\)(do 1 > 0)\(\Leftrightarrow x< 1.\)

Kết hợp ĐKXĐ, A < 1 khi \(x< 1\)và \(x\ne-1;0.\)

c) \(A\inℤ\Leftrightarrow\frac{x}{x-1}\inℤ.\)Mà \(x\inℤ\)\(\Rightarrow x⋮\left(x-1\right)\Rightarrow\left(x-1+1\right)⋮\left(x-1\right)\Rightarrow1⋮\left(x-1\right)\Rightarrow\left(x-1\right)\inƯ\left(1\right)=\left\{1;-1\right\}.\)Ta lập bảng sau:

Vậy để A nguyên thì x = 2.

1111111

\(\text{a) ĐKXĐ: }a\ne1\)
\(\text{b) }M=\frac{a^2+1+a}{a^2+1}:\left[\frac{1}{a-1}-\frac{2a}{a^2\left(a-1\right)+\left(a-1\right)}\right]\)
\(M=\frac{a^2+a+1}{a^2+1}:\left[\frac{1}{a-1}-\frac{2a}{\left(a-1\right)\left(a^2+1\right)}\right]\)
\(M=\frac{a^2+a+1}{a^2+1}:\frac{a^2+1-2a}{\left(a-1\right)\left(a^2+1\right)}\)
\(M=\frac{a^2+a+1}{a^2+1}.\frac{\left(a-1\right)\left(a^2+1\right)}{\left(a-1\right)^2}\)
\(M=\frac{a^2+a+1}{a-1}\)