Bài làm:
PT:
đkxđ: \(x\ne0;x\ne2\)
Ta có: \(\frac{x+2}{x-2}=\frac{2}{x^2-2x}+\frac{1}{x}\)
\(\Leftrightarrow\frac{x\left(x+2\right)}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}+\frac{x-2}{x\left(x-2\right)}\)
\(\Rightarrow x^2+2x=2+x-2\)
\(\Leftrightarrow x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\left(vl\right)\\x+1=0\end{cases}}\Rightarrow x=-1\)
BPT:
Ta có: \(\frac{x+1}{2}-x\le\frac{1}{2}\)
\(\Leftrightarrow\frac{x+1}{2}-x-\frac{1}{2}\le0\)
\(\Leftrightarrow\frac{x+1-2x-1}{2}\le0\)
\(\Leftrightarrow\frac{-x}{2}\le0\)
\(\Rightarrow-x\le0\)
\(\Rightarrow x\ge0\)
a) \(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne2\end{cases}}\)
\(\frac{x+2}{x-2}=\frac{2}{x^2-2x}+\frac{1}{x}\)
\(\Leftrightarrow\frac{2}{x\left(x-2\right)}+\frac{1}{x}-\frac{x+2}{x-2}=0\)
\(\Leftrightarrow\frac{2+x-2-x^2-2x}{x\left(x-2\right)}=0\)
\(\Leftrightarrow-x^2-x=0\)
\(\Leftrightarrow-x\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\left(ktm\right)\\x=-1\left(tm\right)\end{cases}}}\)
Vậy \(S=\left\{-1\right\}\)
b) \(\frac{x+1}{2}-x\le\frac{1}{2}\)
\(\Leftrightarrow x+1-2x-1\le0\)
\(\Leftrightarrow-x\le0\)
\(\Leftrightarrow x\ge0\)
Vậy \(x\ge0\)
ĐKXĐ : \(x\ne0;2\)
\(\frac{x+2}{x-2}=\frac{2}{x^2-2x}+\frac{1}{x}\)
\(\Leftrightarrow\frac{x\left(x+2\right)}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}+\frac{x-2}{x\left(x-2\right)}\)
\(\Leftrightarrow x^2+2x=2+x-2\)
\(\Leftrightarrow x^2+x=0\Leftrightarrow x\left(x+1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\left(ktm\right)\\x=-1\left(tm\right)\end{cases}}\)
\(\frac{x+2}{x-2}=\frac{2}{x^2-2x}+\frac{1}{x}\)( ĐKXĐ \(x\ne0;x\ne2\))
\(\Leftrightarrow\frac{x\left(x+2\right)}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}+\frac{\left(x-2\right)}{x\left(x-2\right)}\)
\(\Leftrightarrow x^2+2x=2+x-2\)
\(\Leftrightarrow x^2+2x-2-x+2=0\)
\(\Leftrightarrow x^2-x=0\)
\(\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\left(ktmđk\right)\\x=1\end{cases}}\)
Vậy S = { 1 }
\(\frac{x+1}{2}-x\le\frac{1}{2}\)
\(\Leftrightarrow\frac{x+1}{2}-\frac{2x}{2}\le\frac{1}{2}\)
\(\Leftrightarrow x+1-2x\le1\)
\(\Leftrightarrow-x\le0\)
\(\Leftrightarrow x\ge0\)
\(\frac{x+2}{x-2}=\frac{2}{x^2-2x}+\frac{1}{x}\left(đk:x\ne0;2\right)\)
\(< =>\frac{x+2}{x-2}=\frac{2}{x\left(x-2\right)}+\frac{1}{x}\)
\(< =>\frac{x\left(x+2\right)}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}+\frac{x-2}{x\left(x-2\right)}\)
\(< =>x\left(x+2\right)=2+x-2\)
\(< =>x^2+2x=x\)\(< =>x^2+x=0\)
\(< =>x\left(x+1\right)=0< =>\orbr{\begin{cases}x=0\left(ktm\right)\\x=-1\left(tm\right)\end{cases}}\)
