a) Ta có: \(B=\left(\frac{21}{x^2-9}+\frac{x-4}{3-x}-\frac{x-1}{3+x}\right)\cdot\left(1-\frac{1}{x+3}\right)\)
\(=\left(\frac{21}{\left(x-3\right)\left(x+3\right)}-\frac{\left(x-4\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{\left(x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\right)\cdot\frac{x+3-1}{x+3}\)
\(=\frac{21-x^2+x+12-x^2+4x-3}{\left(x+3\right)\left(x-3\right)}\cdot\frac{x+2}{x+3}\)
\(=\frac{-2x^2+5x+30}{\left(x-3\right)\cdot\left(x+3\right)}\cdot\frac{x+2}{x+3}\)
\(=\frac{-2x^3+x^2+40x+60}{x^3+3x^2-9x-27}\)
b) ĐKXĐ: \(x\notin\left\{3;-3\right\}\)
Ta có: |2x+1|=5
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=5\\2x+1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=4\\2x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=-3\left(loại\right)\end{matrix}\right.\)
Thay x=2 vào biểu thức \(B=\frac{-2x^3+x^2+40x+60}{x^3+3x^2-9x-27}\), ta được:
\(B=\frac{-2\cdot2^3+2^2+40\cdot2+60}{2^3+3\cdot2^2-9\cdot2-27}=\frac{128}{-25}=\frac{-128}{25}\)
Vậy: \(-\frac{128}{25}\) là giá trị của biểu thức \(B=\frac{-2x^3+x^2+40x+60}{x^3+3x^2-9x-27}\) tại x=2
c) Để \(B=-\frac{3}{5}\) thì \(\frac{-2x^3+x^2+40x+60}{x^3+3x^2-9x-27}=\frac{-3}{5}\)
\(\Leftrightarrow-3\cdot\left(x^3+3x^2-9x-27\right)=5\left(-2x^3+x^2+40x+60\right)\)
\(\Leftrightarrow-3x^3-9x^2+27x+81=-10x^3+5x^2+200x+300\)
\(\Leftrightarrow-3x^3-9x^2+27x+81+10x^3-5x^2-200x-300=0\)
\(\Leftrightarrow7x^3-14x^2-173x-219=0\)
\(\Leftrightarrow x^3-2x^2-\frac{173}{7}x-\frac{219}{7}=0\)
