bài 24:
a: \(A=\dfrac{x}{x+5}-\dfrac{7x-15}{25-x^2}+\dfrac{3}{x-5}\)
\(=\dfrac{x}{x+5}+\dfrac{7x-15}{\left(x-5\right)\left(x+5\right)}+\dfrac{3}{x-5}\)
\(=\dfrac{x\left(x-5\right)+7x-15+3x+15}{\left(x+5\right)\cdot\left(x-5\right)}\)
\(=\dfrac{x^2-5x+10x}{\left(x+5\right)\left(x-5\right)}=\dfrac{x^2+5x}{\left(x+5\right)\left(x-5\right)}=\dfrac{x}{x-5}\)
b: |2x-1|=9
=>\(\left[{}\begin{matrix}2x-1=9\\2x-1=-9\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=5\left(loại\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)
Thay x=-4 vào A, ta được:
\(A=\dfrac{-4}{-5-4}=\dfrac{-4}{-9}=\dfrac{4}{9}\)
Bài 25:
a:
ĐKXĐ: \(x\notin\left\{3;-3;-\dfrac{1}{2}\right\}\)
\(A=\left(\dfrac{x-1}{x+3}+\dfrac{2}{x-3}+\dfrac{x^2+3}{9-x^2}\right):\dfrac{-2}{2x+1}\)
\(=\left(\dfrac{x-1}{x+3}+\dfrac{2}{x-3}-\dfrac{x^2+3}{\left(x-3\right)\left(x+3\right)}\right)\cdot\dfrac{2x+1}{-2}\)
\(=\dfrac{\left(x-1\right)\left(x-3\right)+2\left(x+3\right)-x^2-3}{\left(x+3\right)\cdot\left(x-3\right)}\cdot\dfrac{2x+1}{-2}\)
\(=\dfrac{x^2-4x+3+2x+6-x^2-3}{\left(x+3\right)\left(x-3\right)}\cdot\dfrac{2x+1}{-2}\)
\(=\dfrac{-2x+6}{-2\left(x-3\right)}\cdot\dfrac{2x+1}{x+3}=\dfrac{2x+1}{x+3}\)
b: |x-2|=1
=>\(\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=3\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)
Thay x=1 vào A, ta được:
\(A=\dfrac{2\cdot1+1}{1+3}=\dfrac{3}{4}\)
c: Để A>0 thì \(\dfrac{2x+1}{x+3}>0\)
TH1: \(\left\{{}\begin{matrix}2x+1>0\\x+3>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>-\dfrac{1}{2}\\x>-3\end{matrix}\right.\)
=>\(x>-\dfrac{1}{2}\)
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}x>-\dfrac{1}{2}\\x\ne3\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}2x+1< 0\\x+3< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< -\dfrac{1}{2}\\x< -3\end{matrix}\right.\)
=>x<-3
d: \(A=\dfrac{2x+1}{x+3}=\dfrac{2x+6-5}{x+3}=2-\dfrac{5}{x+3}\)
Để A là số nguyên nhỏ nhất thì \(\dfrac{5}{x+3}\) lớn nhất
=>x+3=1
=>x=-2(nhận)
