Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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.
\(\frac{3x-47}{3x-1}-\frac{4x-47}{2x-1}>0\)
\(\Leftrightarrow\frac{\left(3x-47\right)\left(2x-1\right)-\left(4x-47\right)\left(3x-1\right)}{\left(3x-1\right)\left(2x-1\right)}>0\)
\(\Leftrightarrow\frac{-6x\left(x-8\right)}{\left(3x-1\right)\left(2x-1\right)}>0\Rightarrow\left[{}\begin{matrix}0< x< \frac{1}{3}\\\frac{1}{2}< x< 8\end{matrix}\right.\)
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
a/ ĐKXĐ: \(x\ne\pm5\)
\(\Leftrightarrow\left(x+5\right)^2-\left(x-5\right)^2=20\)
\(\Leftrightarrow\left(x^2+10x+25\right)-\left(x^2-10x+25\right)=20\)
\(\Leftrightarrow20x=20\Rightarrow x=1\)
b/ ĐKXĐ: \(x\ne\pm4\)
\(\Leftrightarrow2x\left(x-4\right)-4x=0\)
\(\Leftrightarrow2x^2-12x=0\)
\(\Leftrightarrow2x\left(x-6\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
c/ ĐKXĐ: \(x\ne\pm\frac{2}{3}\)
\(\Leftrightarrow\left(3x+2\right)^2-6\left(3x-2\right)=9x^2\)
\(\Leftrightarrow9x^2+12x+4-18x+12=9x^2\)
\(\Leftrightarrow6x=16\Rightarrow x=\frac{8}{3}\)