a) ĐKXĐ: x \(\ne\) \(\pm\) 1
Ta có:
\(P=\left(\frac{x}{1-x^2}-\frac{6\left(1+x\right)}{\left(1-x\right)\left(1+x\right)}+\frac{3\left(1-x\right)}{\left(1-x\right)\left(1+x\right)}\right).\frac{x^2-1}{4x^2+1}\)
\(P=\left(\frac{x-\left(6+6x\right)+\left(3-3x\right)}{1-x^2}\right).\frac{x^2-1}{4x^2+1}\)
\(P=\frac{8x+3}{x^2-1}.\frac{x^2-1}{4x^2+1}=\frac{8x+3}{4x^2+1}\)
b) Để P = 3 thì 8x + 3 = 12x2 + 3
\(\Leftrightarrow8x=12x^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{2}{3}\end{matrix}\right.\left(t/m\right)\)
c) Dễ thấy 4x2 + 1 \(\ge\) 0 với mọi x. Do đó để P < 0 thì x2 - 1 < 0
\(\Leftrightarrow\) x2 < 1
\(\Leftrightarrow\) -1 < x < 1 (t/m)
a/ Ta có :
\(P=\left(\frac{x}{1-x^2}-\frac{6}{1-x}+\frac{3}{1+x}\right):\frac{4x^2+1}{x^2-1}\) \(\left(x\ne\pm1\right)\)
\(=\left(\frac{x}{\left(1-x\right)\left(1+x\right)}-\frac{6}{1-x}+\frac{3}{1+x}\right):\frac{4x^2+1}{\left(x-1\right)\left(x+1\right)}\)
\(\Leftrightarrow\frac{x-6\left(1+x\right)+3\left(1-x\right)}{\left(1-x\right)\left(1+x\right)}.\frac{\left(x-1\right)\left(x+1\right)}{4x^2+1}\)
\(=\frac{x-6-6x+3-3x}{\left(1-x\right)\left(1+x\right)}.\frac{\left(x+1\right)\left(x-1\right)}{4x^2+1}\)
\(=\frac{-8x-3}{\left(1+x\right)\left(1-x\right)}.\frac{\left(x+1\right)\left(x-1\right)}{4x^2-1}\)
\(=\frac{8x+3}{4x^2+1}\)
b/ \(P=3\Leftrightarrow\frac{8x+3}{4x^2+1}=3\)
\(\Leftrightarrow8x+3=12x^2+3\)
\(\Leftrightarrow12x^2-8x=0\)
\(\Leftrightarrow4x\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=0\\3x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{2}{3}\end{matrix}\right.\)
c/ \(P< 0\)
\(\Leftrightarrow\frac{8x+3}{4x^2+1}< 0\)
Mà \(4x^2+1>0\forall x\)
\(\Leftrightarrow8x+3< 0\Leftrightarrow x< -\frac{3}{8}\)
Vậy....
