\(ĐKXĐ:\left\{{}\begin{matrix}x\ne0\\x\ne1\\x\ne2\end{matrix}\right.\)

\(\dfrac{1}{x-2}+\dfrac{1}{x-1}>\dfrac{1}{x}\\ \Leftrightarrow\dfrac{x-1+x-2}{\left(x-1\right)\left(x-2\right)}>\dfrac{1}{x}\\ \Leftrightarrow\dfrac{2x-3}{x^2-3x+2}>\dfrac{1}{x}\\ \Leftrightarrow x\left(2x-3\right)>x^2-3x+2\\ \Leftrightarrow2x^2-3x>x^2-3x+2\\ \Leftrightarrow x^2>2\\ \Leftrightarrow\left[{}\begin{matrix}x>\sqrt{2}\\x< -\sqrt{2}\end{matrix}\right.\)

ĐKXĐ:\(x\ne-1\)

\(\dfrac{x-1}{x+1}\le5+x\\ \Leftrightarrow x-1\le\left(x+1\right)\left(5+x\right)\\ \Leftrightarrow x-1\le x^2+6x+5\\ \Leftrightarrow x^2+5x+6\ge0\\ \Leftrightarrow\left[{}\begin{matrix}x\le-3\\x\ge-2\end{matrix}\right.\)

ĐKXĐ:\(\left\{{}\begin{matrix}x\ne-2\\x\ne3\end{matrix}\right.\)

\(\dfrac{1}{x+2}< \dfrac{3}{x-3}\\ \Leftrightarrow x-3< 3\left(x+2\right)\\ \Leftrightarrow x-3< 3x+6\\ \Leftrightarrow2x+9< 0\\ \Rightarrow x< -\dfrac{9}{2}\)

\(ĐKXĐ:\left\{{}\begin{matrix}x\ne-1\\x\ne4\end{matrix}\right.\)

\(\dfrac{14x}{x+1}< \dfrac{9x-30}{x-4}\\ \Leftrightarrow14x\left(x-4\right)< \left(9x-30\right)\left(x+1\right)\\ \Leftrightarrow14x^2-56x< 9x^2-21x-30\\ \Leftrightarrow5x^2-35x+30< 0\\ \Leftrightarrow1< x< 6\)

ĐKXĐ: \(1< x< 9\)

Đặt \(\left\{{}\begin{matrix}\sqrt{9-x}=a\\\sqrt{x-1}=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a;b>0\\a^2+b^2=8\end{matrix}\right.\) \(\Rightarrow\left(a+b\right)^2\le16\Rightarrow a+b\le4\)

\(BPT\Leftrightarrow\dfrac{a^2-1}{a}+\dfrac{b^2-1}{b}\ge3\) (1)

Đặt \(P=\dfrac{a^2-1}{a}+\dfrac{b^2-1}{b}-3\)

\(P=a+b-\left(\dfrac{1}{a}+\dfrac{1}{b}\right)-3\le a+b-\dfrac{4}{a+b}-3\)

\(P\le\dfrac{\left(a+b\right)^2-3\left(a+b\right)-4}{a+b}=\dfrac{\left(a+b+1\right)\left(a+b-4\right)}{a+b}\le0\)

\(\Rightarrow\dfrac{a^2-1}{a}+\dfrac{b^2-1}{b}\le3\) (2)

(1); (2) \(\Rightarrow\dfrac{a^2-1}{a}+\dfrac{b^2-1}{b}=3\)

Dấu "=" xảy ra khi và chỉ khi: \(a=b=2\Leftrightarrow x=5\)

Vậy BPT đã cho có nghiệm duy nhất \(x=5\)