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ĐKXĐ : \(x\ne\pm2;x\ne0;x\ne3\)
\(A=\left(\frac{4x}{2+x}+\frac{8x^2}{4-x^2}\right):\left(\frac{x-1}{x^2-2x}-\frac{2}{x}\right)\)
\(=\frac{4x\left(2-x\right)+8x^2}{\left(2-x\right)\left(2+x\right)}:\frac{x-1-2\left(x-2\right)}{x\left(x-2\right)}\)
\(=\frac{8x-4x^2+8x^2}{\left(2-x\right)\left(2+x\right)}:\frac{x-1-2x+4}{x\left(x-2\right)}\)
\(=\frac{8x+4x^2}{\left(2-x\right)\left(2+x\right)}:\frac{3-x}{x\left(x-2\right)}\)
\(=\frac{8x+4x^2}{\left(2-x\right)\left(2+x\right)}.\frac{x\left(x-2\right)}{3-x}\) \(=\frac{4x\left(2+x\right)}{\left(2-x\right)\left(2+x\right)}.\frac{x\left(2-x\right)}{x-3}\)
\(=\frac{4x^2}{x-3}\)
\(A< 0\Leftrightarrow\frac{4x^2}{x-3}< 0\Leftrightarrow x-3< 0\) ( do \(4x^2>0\) )
\(\Leftrightarrow x< 3\)
Vậy :........
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.
Bài 2:
a) ĐK: $x\geq \pm \frac{1}{2}; x\neq 0$
\(\left(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right):\frac{4x}{10x-5}=\frac{(2x+1)^2-(2x-1)^2}{(2x-1)(2x+1)}.\frac{10x-5}{4x}\)
\(\frac{4x^2+4x+1-(4x^2-4x+1)}{(2x-1)(2x+1)}.\frac{5(2x-1)}{4x}=\frac{8x}{(2x-1)(2x+1)}.\frac{5(2x-1)}{4x}\)
\(=\frac{10}{2x+1}\)
b) ĐK : $x\neq 0;-1$
\(\left(\frac{1}{x^2+x}-\frac{2-x}{x+1}\right):\left(\frac{1}{x}+x-2\right)=\left(\frac{1}{x(x+1)}-\frac{x(2-x)}{x(x+1)}\right):\frac{1+x^2-2x}{x}\)
\(=\frac{1-2x+x^2}{x(x+1)}.\frac{x}{1+x^2-2x}=\frac{x}{x(x+1)}=\frac{1}{x+1}\)
Bài 3:
a) ĐKXĐ: \(x\neq \pm 1\)
b)
\(A=\left(\frac{x+1}{2x-2}-\frac{3}{1-x^2}-\frac{x+3}{2x+2}\right).\frac{4x^2-4}{5}\)
\(=\left[\frac{(x+1)^2}{2(x-1)(x+1)}+\frac{6}{2(x-1)(x+1)}-\frac{(x+3)(x-1)}{2(x+1)(x-1)}\right].\frac{4(x^2-1)}{5}\)
\(=\frac{(x+1)^2+6-(x^2+2x-3)}{2(x-1)(x+1)}.\frac{4(x-1)(x+1)}{5}\)
\(=\frac{10}{2(x-1)(x+1)}.\frac{4(x-1)(x+1)}{5}=4\)
c ơn nghen