a) Ta có: \(A=\left(\dfrac{2x}{2x^2-5x+3}-\dfrac{5}{2x-3}\right):\left(3+\dfrac{2}{1-x}\right)\)
\(=\dfrac{2x-5\left(x-1\right)}{\left(2x-3\right)\left(x-1\right)}:\dfrac{3\left(x-1\right)-2}{x-1}\)
\(=\dfrac{2x-5x+5}{2x-3}\cdot\dfrac{1}{3x-3-2}\)
\(=\dfrac{-3x+5}{2x-3}\cdot\dfrac{1}{3x-5}\)
\(=\dfrac{-1}{2x-3}\)
c) Để A>0 thì 2x-3<0
hay \(x< \dfrac{3}{2}\)
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}x< \dfrac{3}{2}\\x\ne1\end{matrix}\right.\)
Tham khảo:Cho biểu thức P= \((\frac{2x}{2x^2-5x+3}-\frac{5}{2x-3}):(3+\frac{2}{1-x})\) a) Rút gọn P b) Tính P với |3x-2|+1=5 c)... - Hoc24
a) ĐKXĐ:
Ta có:
a, \(A=\left(\dfrac{2x}{2x^2-5x+3}-\dfrac{5}{2x-3}\right):\left(3+\dfrac{2}{1-x}\right)\)ĐK : \(x\ne1;\dfrac{3}{2}\)
\(=\left(\dfrac{2x-5\left(x-1\right)}{\left(2x-3\right)\left(x-1\right)}\right):\left(\dfrac{3-x+2}{1-x}\right)=\left(\dfrac{5-3x}{\left(2x-3\right)\left(x-1\right)}\right):\left(\dfrac{5-x}{1-x}\right)\)
\(=\dfrac{3x-5}{\left(2x-3\right)\left(5-x\right)}\)
b, \(\left|3x-2\right|+1=5\Leftrightarrow\left[{}\begin{matrix}3x-2=4\\3x-2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{2}{3}\end{matrix}\right.\)
Với x = 2 \(A=\dfrac{6-5}{1.3}=\dfrac{1}{3}\)
Với x = -2/3 \(A=\dfrac{3\left(-\dfrac{2}{3}\right)-5}{\left(-\dfrac{2}{3}-3\right)\left(5+\dfrac{2}{3}\right)}=\dfrac{-2-5}{-\dfrac{11}{3}.\dfrac{17}{3}}=-\dfrac{7}{-\dfrac{187}{9}}=\dfrac{63}{187}\)
