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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) \(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne2\end{cases}}\)
\(Q=\left(\frac{2x-x^2}{2x^2+8}-\frac{2x^2}{x^3-2x^2+4x-8}\right).\left(\frac{2}{x^2}+\frac{1-x}{x}\right)\)
\(\Leftrightarrow Q=\left(\frac{x\left(2-x\right)}{2\left(x^2+4\right)}-\frac{2x^2}{\left(x-2\right)\left(x^2+4\right)}\right).\frac{2+x\left(1-x\right)}{x^2}\)
\(\Leftrightarrow Q=\frac{-x\left(x-2\right)^2-4x^2}{2\left(x-2\right)\left(x^2+4\right)}.\frac{2+x-x^2}{x^2}\)
\(\Leftrightarrow Q=\frac{x\left(x^2-4x+4\right)-4x^2}{2\left(x-2\right)\left(x^2+4\right)}.\frac{\left(x-2\right)\left(x+1\right)}{x^2}\)
\(\Leftrightarrow Q=\frac{x\left(x^2+4\right)}{2\left(x^2+4\right)}.\frac{x+1}{x^2}\)
\(\Leftrightarrow Q=\frac{x+1}{2x}\)
b) Để \(Q\inℤ\)
\(\Leftrightarrow x+1⋮2x\)
\(\Leftrightarrow2\left(x+1\right)⋮2x\)
\(\Leftrightarrow2x+2⋮2x\)
\(\Leftrightarrow2⋮2x\)
\(\Leftrightarrow2x\inƯ\left(2\right)\)
\(\Leftrightarrow2x\in\left\{\pm1;\pm2\right\}\)
\(\Leftrightarrow x\in\left\{\pm\frac{1}{2};\pm1\right\}\)
Mà \(x\inℤ\)
Vậy để \(Q\inℤ\Leftrightarrow x\in\left\{1;-1\right\}\)