Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
ĐKXĐ:\(x\ne\pm2;x\ne-3;x\ne0\)
\(P=1+\frac{x-3}{x^2+5x+6}\left(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3x^2-12}-\frac{1}{x+2}\right)\)
\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\left[\frac{8x^2}{4x^2\left(x-2\right)}-\frac{3x}{3\left(x^2-4\right)}-\frac{1}{x+2}\right]\)
\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\left(\frac{2}{x-2}-\frac{x}{x^2-4}-\frac{1}{x+2}\right)\)
\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\left[\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right]\)
\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\cdot\frac{2x+4-x-x+4}{\left(x-2\right)\left(x+2\right)}\)
\(=1+\frac{8\left(x-3\right)}{\left(x+2\right)^2\left(x+3\right)\left(x-2\right)}\)
Đề sai à ??
\(\frac{3x-7}{5}=\frac{2x-1}{3}\)
\(\Leftrightarrow9x-21=10x-5\)
\(\Leftrightarrow-x=16\Leftrightarrow x=-16\)
\(\frac{4x-7}{12}-x=\frac{3x}{8}\)
\(\Leftrightarrow\frac{4x-7-12x}{12}=\frac{3x}{8}\)
\(\Leftrightarrow\frac{-7-8x}{12}=\frac{3x}{8}\)
\(\Leftrightarrow-56-64x=36x\)
\(\Leftrightarrow-56=100x\Leftrightarrow x=\frac{-14}{25}\)
\(\frac{x-2009}{1234}+\frac{x-2009}{5678}-\frac{x-2009}{197}=0\)
\(\Leftrightarrow\left(x-2019\right)\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)=0\)
Vì \(\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)\ne0\)nên x - 2019 = 0
Vậy x = 2019
\(\frac{5x-8}{3}=\frac{1-3x}{2}\)
\(\Leftrightarrow10x-16=3-9x\)
\(\Leftrightarrow19x=19\Leftrightarrow x=1\)
a) \(\frac{x-2}{3x+2}=0\Rightarrow x-2=0\Rightarrow x=2\)
b) \(\frac{x+8}{x+9}>0\)
TH1 : \(\hept{\begin{cases}x+8< 0\\x+9< 0\end{cases}\Rightarrow\hept{\begin{cases}x< -8\\x< -9\end{cases}}\Rightarrow x< -9}\)
TH2 : \(\hept{\begin{cases}x+8>0\\x+9>0\end{cases}\Rightarrow\hept{\begin{cases}x>-8\\x>-9\end{cases}\Rightarrow}x>-8}\)
Vậy khi x < -9 hoặc x > - 8 thì \(\frac{x+8}{x+9}>0\)
c) \(\frac{x-2}{x-6}< 0\)
TH1 : \(\hept{\begin{cases}x-2< 0\\x-6>0\end{cases}}\Rightarrow\hept{\begin{cases}x< 2\\x>6\end{cases}}\Rightarrow x\in\varnothing\)
TH2 : \(\hept{\begin{cases}x-2>0\\x-6< 0\end{cases}}\Rightarrow\hept{\begin{cases}x>2\\x< 6\end{cases}}\Rightarrow2< x< 6\)
Vậy khi 2 < x < 6 thì \(\frac{x-2}{x-6}< 0\)
a)\(\frac{x-2}{3x+2}=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\3x+2=0\left(vl\right)\end{cases}}\Leftrightarrow x=2\)
vậy x=2 thì \(\frac{x-2}{3x+2}=0\)
b)\(\frac{x+8}{x+9}>0\)
=> x+8 và x+9 cùng dấu
\(th1\orbr{\begin{cases}x+8>0\\x+9>0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x>-8\\x>-9\end{cases}}\Leftrightarrow x>-8\left(1\right)\)
\(th2\orbr{\begin{cases}x+8< 0\\x+9< 0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x< -8\\x< -9\end{cases}}\Leftrightarrow x< -9\left(2\right)\)
từ (1) và (2) =>\(-8< x< -9\)
\(\Rightarrow x=-7\)
vậy với x=-7 thì\(\frac{x+8}{x+9}>0\)
c) \(\frac{x-2}{x-6}< 0\)
=> x-2 và x-6 khác dấu
\(th1\hept{\begin{cases}x-2>0\\x-6< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x>2\\x< 6\end{cases}}}\Leftrightarrow2< x< 6\left(tm\right)\)
\(th2\hept{\begin{cases}x-2< 0\\x-6>0\end{cases}\Leftrightarrow\hept{\begin{cases}x< 2\\x>6\end{cases}}}\Leftrightarrow6< x< 2\left(ktm\right)\)
từ \(2< x< 6\Rightarrow x\in\left\{3,4,5\right\}\)
vậy với \(x\in\left\{3,4,5\right\}\)thì \(\frac{x-2}{x-6}< 0\)