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\(ĐKXĐ:\left\{{}\begin{matrix}x\ne0\\x\ne1\\x\ne2\end{matrix}\right.\)
\(\dfrac{1}{x-2}+\dfrac{1}{x-1}>\dfrac{1}{x}\\ \Leftrightarrow\dfrac{x-1+x-2}{\left(x-1\right)\left(x-2\right)}>\dfrac{1}{x}\\ \Leftrightarrow\dfrac{2x-3}{x^2-3x+2}>\dfrac{1}{x}\\ \Leftrightarrow x\left(2x-3\right)>x^2-3x+2\\ \Leftrightarrow2x^2-3x>x^2-3x+2\\ \Leftrightarrow x^2>2\\ \Leftrightarrow\left[{}\begin{matrix}x>\sqrt{2}\\x< -\sqrt{2}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ:\(\left\{{}\begin{matrix}x\ne1\\x\ne2\\x\ne7\end{matrix}\right.\)
\(\dfrac{2\left(x-4\right)}{\left(x-1\right)\left(x-7\right)}\ge\dfrac{1}{x-2}\\ \Leftrightarrow\dfrac{2x-8}{x^2-8x+7}\ge\dfrac{1}{x-2}\\ \Leftrightarrow\left(2x-8\right)\left(x-2\right)\ge x^2-8x+7\)
\(\Leftrightarrow2x^2-12x+16\ge x^2-8x+7\\ \Leftrightarrow x^2-4x+9\ge0\left(luôn.đúng\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
a: =>2x^2-4x+1=x^2+x+5
=>x^2-5x-4=0
=>\(x=\dfrac{5\pm\sqrt{41}}{2}\)
b: =>11x^2-14x-12=3x^2+4x-7
=>8x^2-18x-5=0
=>x=5/2 hoặc x=-1/4
\(ĐKXĐ:\left\{{}\begin{matrix}x\ne-1\\x\ne4\end{matrix}\right.\)
\(\dfrac{14x}{x+1}< \dfrac{9x-30}{x-4}\\ \Leftrightarrow14x\left(x-4\right)< \left(9x-30\right)\left(x+1\right)\\ \Leftrightarrow14x^2-56x< 9x^2-21x-30\\ \Leftrightarrow5x^2-35x+30< 0\\ \Leftrightarrow1< x< 6\)