a) \(A=\frac{x-3}{4x-8}\left(ĐKXĐ:x\ne2\right).\)
Ta có:
\(\left|2x-1\right|=3\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=4\\2x=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4:2\\x=\left(-2\right):2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\left(KTM\right)\\x=-1\left(TM\right)\end{matrix}\right.\)
+ Thay \(x=-1\) vào biểu thức A ta được:
\(A=\frac{\left(-1\right)-3}{4.\left(-1\right)-8}\)
\(\Rightarrow A=\frac{-4}{\left(-4\right)-8}\)
\(\Rightarrow A=\frac{-4}{-12}\)
\(\Rightarrow A=\frac{1}{3}.\)
Vậy giá trị của biểu thức A tại \(x=-1\) là: \(\frac{1}{3}.\)
b) Rút gọn B:
\(B=\frac{3}{x+2}-\frac{x}{x-2}+\frac{x^2+3}{x^2-4}\)
\(B=\frac{3}{x+2}-\frac{x}{x-2}+\frac{x^2+3}{\left(x-2\right).\left(x+2\right)}\)
\(B=\frac{3.\left(x-2\right)}{\left(x-2\right).\left(x+2\right)}-\frac{x.\left(x+2\right)}{\left(x-2\right).\left(x+2\right)}+\frac{x^2+3}{\left(x-2\right).\left(x+2\right)}\)
\(B=\frac{3x-6}{\left(x-2\right).\left(x+2\right)}-\frac{x^2+2x}{\left(x-2\right).\left(x+2\right)}+\frac{x^2+3}{\left(x-2\right).\left(x+2\right)}\)
\(B=\frac{3x-6}{\left(x-2\right).\left(x+2\right)}+\frac{-\left(x^2+2x\right)}{\left(x-2\right).\left(x+2\right)}+\frac{x^2+3}{\left(x-2\right).\left(x+2\right)}\)
\(B=\frac{3x-6-x^2-2x+x^2+3}{\left(x-2\right).\left(x+2\right)}\)
\(B=\frac{x-3}{\left(x-2\right).\left(x+2\right)}.\)
Chúc bạn học tốt!
