a, ĐKXĐ : x khác -1 và 1

b, A = 2x^2+4x+2/(x-1).(x+1)  .  (x-1)/10

  = 2.(x^2+2x+1)/10.(x+1)

  = (x+1)^2/5.(x+1)

  = x+1/5

k mk nha

a, ĐKXĐ: \(x\ne\pm1\)

b, \(A=\left(\frac{2x}{x-1}+\frac{4x}{x^2-1}-\frac{2}{x+1}\right)\frac{x-1}{10}\)

\(A=\left(\frac{2x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{4x}{\left(x-1\right)\left(x+1\right)}-\frac{2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\right)\frac{x-1}{10}\)

\(A=\frac{2x^2+2x+4x-2x+2}{\left(x-1\right)\left(x+1\right)}.\frac{x-1}{10}\)

\(A=\frac{2x^2+4x+2}{10\left(x+1\right)}\)

\(A=\frac{2\left(x+1\right)^2}{10\left(x+1\right)}\)

\(A=\frac{\left(x+1\right)}{5}\)

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 !

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.