a) \(B=\left(\frac{9-3x}{\left(x-1\right)\left(x+5\right)}+\frac{\left(x+5\right)^2}{\left(x-1\right)\left(x+5\right)}-\frac{\left(x+1\right)\left(x-1\right)}{\left(x+5\right)\left(x-1\right)}\right)\)\(:\frac{7\left(x-2\right)}{\left(x-1\right)\left(x+1\right)}\)
\(B=\frac{9-3x+x^2+10x+25-\left(x^2-1\right)}{\left(x-1\right)\left(x+5\right)}\cdot\frac{\left(x-1\right)\left(x+1\right)}{7\left(x-2\right)}\)
\(B=\frac{7x+35}{\left(x-1\right)\left(x+5\right)}\cdot\frac{\left(x-1\right)\left(x+1\right)}{7\left(x-2\right)}\)
\(B=\frac{7\left(x+5\right)\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+5\right)\cdot7\left(x-2\right)}=\frac{x+1}{x-2}\)
b) \(\left(x+5\right)^2-9x-45=0\)
\(\Leftrightarrow x^2+10x+25-9x-45=0\)
\(\Leftrightarrow x^2+x-20=0\)
\(\Leftrightarrow x^2-4x+5x-20=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-5\left(KTM\right)\\x=4\left(TM\right)\end{matrix}\right.\)
Với x = 4 ta có \(B=\frac{4+1}{4-2}=\frac{5}{2}\)
c) \(B=\frac{x+1}{x-2}\) đạt giá trị nguyên
\(\Leftrightarrow x+1⋮x-2\)
\(\Leftrightarrow x-2+3⋮x-2\)
\(\Leftrightarrow3⋮x-2\) \(\Leftrightarrow\left(x-2\right)\in\left\{-3;-1;1;3\right\}\)
\(\Leftrightarrow x\in\left\{-1;1;3;5\right\}\)
d) \(B=-\frac{3}{4}\Leftrightarrow\frac{x+1}{x-2}=\frac{-3}{4}\)
\(\Leftrightarrow4\left(x+1\right)=-3\left(x-2\right)\)
\(\Leftrightarrow4x+4=-3x+6\)
\(\Leftrightarrow7x=2\Leftrightarrow x=\frac{2}{7}\)