Question

1) \(\frac{x^2+2x+5}{x+4}\ge x-3\)

2) \(\frac{x^2-3x-1}{2-x}>-x\)

3) \(\frac{3x-47}{3x-1}>\frac{4x-47}{2x-1}\)

4) \(x+\frac{9}{x+2}\ge4\)

5) \(\frac{\left(x-1\right)^3\left(x+2\right)^4\left(x+6\right)}{\left(x-7\right)^3\left(x-2\right)^2}\le0\)

6) \(x^4\ge\left(x^2+4x+2\right)^27x^2-7x+10< 0\)

Answer 1

1.

\(\frac{x^2+2x+5}{x+4}-\left(x-3\right)\ge0\)

\(\Leftrightarrow\frac{x^2+2x+5-\left(x-3\right)\left(x+4\right)}{x+4}\ge0\)

\(\Leftrightarrow\frac{x+17}{x+4}\ge0\Rightarrow\left[{}\begin{matrix}x>-4\\x\le-12\end{matrix}\right.\)

2.

\(\frac{x^2-3x-1}{2-x}+x>0\)

\(\Leftrightarrow\frac{x^2-3x-1+x\left(2-x\right)}{2-x}>0\)

\(\Leftrightarrow\frac{-x-1}{2-x}>0\Rightarrow\left[{}\begin{matrix}x< -1\\x>2\end{matrix}\right.\)

3.

\(\frac{3x-47}{3x-1}-\frac{4x-47}{2x-1}>0\)

\(\Leftrightarrow\frac{\left(3x-47\right)\left(2x-1\right)-\left(4x-47\right)\left(3x-1\right)}{\left(3x-1\right)\left(2x-1\right)}>0\)

\(\Leftrightarrow\frac{-6x\left(x-8\right)}{\left(3x-1\right)\left(2x-1\right)}>0\Rightarrow\left[{}\begin{matrix}0< x< \frac{1}{3}\\\frac{1}{2}< x< 8\end{matrix}\right.\)

Answer 2

4.

\(\frac{x\left(x+2\right)+9}{x+2}-4\ge0\)

\(\Leftrightarrow\frac{x^2+2x+9-4\left(x+2\right)}{x+2}\ge0\)

\(\Leftrightarrow\frac{x^2-2x+1}{x+2}\ge0\)

\(\Leftrightarrow\frac{\left(x-1\right)^2}{x+2}\ge0\Rightarrow x>-2\)

5.

\(\frac{\left(x-1\right)^3\left(x+2\right)^4\left(x+6\right)}{\left(x-7\right)^3\left(x-2\right)^2}\le0\Rightarrow\left[{}\begin{matrix}x\le-6\\1\le x< 2\\2< x< 7\\x=-2\end{matrix}\right.\)

6. Xem lại đề