1: Thay \(x=\dfrac{1}{25}\) vào C, ta được:

\(C=\left(\dfrac{1}{5}+2\right):\left(\dfrac{1}{5}-3\right)=\dfrac{11}{5}:\dfrac{-14}{5}=-\dfrac{11}{14}\)

2: Để C=-2 thì \(\sqrt{x}+2=-2\left(\sqrt{x}-3\right)\)

\(\Leftrightarrow\sqrt{x}+2+2\sqrt{x}-6=0\)

\(\Leftrightarrow3\sqrt{x}=4\)

hay \(x=\dfrac{16}{9}\)

Để \(C=\dfrac{7}{5}\) thì \(\dfrac{\sqrt{x}+2}{\sqrt{x}-3}=\dfrac{7}{5}\)

\(\Leftrightarrow7\sqrt{x}-21=2\sqrt{x}+10\)

\(\Leftrightarrow5\sqrt{x}=31\)

hay \(x=\dfrac{961}{25}\)

1) \(Q=-x\) khi:

\(\dfrac{x-3}{x+1}=-x\)

\(\Leftrightarrow x-3=-x\left(x+1\right)\)

\(\Leftrightarrow x-3=-x^2-x\)

\(\Leftrightarrow x-3+x^2+x\)

\(\Leftrightarrow x^2+2x-3=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)

2) \(Q< 1\) khi:

\(\dfrac{x-3}{x+1}< 1\)

\(\Leftrightarrow x-3< x+1\)

\(\Leftrightarrow x-x< 1+3\)

\(\Leftrightarrow0< 4\) (luôn đúng) 

Vậy \(Q< 0\) với mọi x 

3) \(Q=m\) khi:

\(\dfrac{x-3}{x+1}=m\)

\(\Leftrightarrow x-3=m\left(x+1\right)\)

\(\Leftrightarrow x-3=mx+m\)

\(\Leftrightarrow x-mx=m+3\)

\(\Leftrightarrow x\left(1-m\right)=m+3\)

\(\Leftrightarrow1-m\ne0\)

\(\Leftrightarrow m\ne1\)

a: ĐKXĐ: x>0; x<>9

b: \(A=\dfrac{x+3\sqrt{x}+x-3\sqrt{x}}{x-9}:\dfrac{\sqrt{x}+3-3}{\sqrt{x}+3}\)

\(=\dfrac{2x}{x-9}\cdot\dfrac{\sqrt{x}+3}{\sqrt{x}}=\dfrac{2\sqrt{x}}{\sqrt{x}-3}\)

c: Để A=-1 thì 2 căn x=-căn x+3

=>x=1

a, ĐKXĐ : \(\left\{{}\begin{matrix}\dfrac{3x-5}{x-1}\ge0\\x-1\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}3x-5\ge0\\x-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}3x-5\le0\\x-1< 0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge\dfrac{5}{3}\\x>1\end{matrix}\right.\\\left\{{}\begin{matrix}x\le\dfrac{5}{3}\\x< 1\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\ge\dfrac{5}{3}\\x< 1\end{matrix}\right.\)

Vậy ...

b, Ta có : \(A=\sqrt{\dfrac{3x-5}{x-1}}=3\)

\(\Leftrightarrow3x-5=9x-9\)

\(\Leftrightarrow x=\dfrac{2}{3}\left(TM\right)\)

Vậy ...

\(y'=3x^2-2x+m\)

\(y'\ge0\Leftrightarrow\left\{{}\begin{matrix}a>0\\\Delta'\le0\end{matrix}\right.\Leftrightarrow1-3m\le0\Leftrightarrow m\ge\dfrac{1}{3}\)

a: \(\dfrac{1}{x-1}-\dfrac{x^3-x}{x^2+1}\cdot\left(\dfrac{x}{x^2-2x+1}-\dfrac{1}{x^2-1}\right)\)

\(=\dfrac{1}{x-1}-\dfrac{x\left(x-1\right)\left(x+1\right)}{x^2+1}\cdot\left(\dfrac{x}{\left(x-1\right)^2}-\dfrac{1}{\left(x-1\right)\left(x+1\right)}\right)\)

\(=\dfrac{1}{x-1}-\dfrac{x\left(x-1\right)\left(x+1\right)}{x^2+1}\cdot\dfrac{x\left(x+1\right)-x+1}{\left(x-1\right)^2\cdot\left(x+1\right)}\)

\(=\dfrac{1}{x-1}-\dfrac{x}{x^2+1}\cdot\dfrac{x^2+x-x+1}{x-1}\)

\(=\dfrac{1-x}{x-1}=-1\)

b: \(\dfrac{x}{6-x}+\left(\dfrac{x}{\left(x-6\right)\left(x+6\right)}-\dfrac{x-6}{x\left(x+6\right)}\right):\dfrac{2x-6}{x^2+6x}\)

\(=\dfrac{x}{6-x}+\dfrac{x^2-\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}\cdot\dfrac{x\left(x+6\right)}{2\left(x-3\right)}\)

\(=\dfrac{x}{6-x}+\dfrac{x^2-x^2+12x-36}{x-6}\cdot\dfrac{1}{2\left(x-3\right)}\)

\(=\dfrac{x}{6-x}+\dfrac{12\left(x-3\right)}{2\left(x-3\right)\left(x-6\right)}\)

\(=\dfrac{x}{6-x}+\dfrac{6}{x-6}=\dfrac{-x+6}{x-6}=-1\)