\(E=\dfrac{12x-45}{\left(x-3\right)\left(x-4\right)}-\dfrac{\left(x+5\right)\left(x-3\right)}{\left(x-3\right)\left(x-4\right)}-\dfrac{\left(2x+3\right)\left(x-4\right)}{\left(x-3\right)\left(x-4\right)}\)
\(=\dfrac{12x-45}{\left(x-3\right)\left(x-4\right)}-\dfrac{x^2+2x-15}{\left(x-3\right)\left(x-4\right)}-\dfrac{2x^2-5x-12}{\left(x-3\right)\left(x-4\right)}\)
\(=\dfrac{12x-45-x^2-2x+15-2x^2+5x+12}{\left(x-3\right)\left(x-4\right)}\)
\(=\dfrac{-3x^2+15x-18}{\left(x-3\right)\left(x-4\right)}=\dfrac{-3\left(x-3\right)\left(x-2\right)}{\left(x-3\right)\left(x-4\right)}\)
\(=\dfrac{-3x+6}{x-4}\)
b.
\(x^2-5x+6=0\Rightarrow\left[{}\begin{matrix}x=2\\x=3\left(ktm\right)\end{matrix}\right.\)
\(\Rightarrow E=\dfrac{-3.2+6}{2-4}=0\)
c.
\(E=\dfrac{-3x+6}{x-4}=\dfrac{-3x+12-6}{x-4}=-3-\dfrac{6}{x-4}\)
\(E\in Z\Rightarrow\dfrac{6}{x-4}\in Z\Rightarrow x-4=Ư\left(6\right)\)
\(\Rightarrow x-4=\left\{-6;-3;-2;-1;1;2;3;6\right\}\)
\(\Rightarrow x=\left\{-2;1;2;3;5;6;7;10\right\}\)
Thử lại chỉ có \(x=\left\{2;3\right\}\) cho giá trị E là số tự nhiên
d.\(E=\dfrac{-3\left(x-2\right)}{x-4}⋮3\Rightarrow\dfrac{x-2}{x-4}\in Z\Rightarrow\dfrac{x-4+2}{x-4}\in Z\)
\(\Rightarrow1+\dfrac{2}{x-4}\in Z\Rightarrow\dfrac{2}{x-4}\in Z\)
\(\Rightarrow x-4=Ư\left(2\right)=\left\{-2;-1;1;2\right\}\)
\(\Rightarrow x=\left\{2;3;5;6\right\}\)
a: \(=\dfrac{12x-45-x^2-2x+15-\left(2x+3\right)\left(x-4\right)}{\left(x-4\right)\left(x-3\right)}\)
\(=\dfrac{-x^2+10x-30-2x^2+8x-3x+12}{\left(x-4\right)\left(x-3\right)}\)
\(=\dfrac{-3x^2+15x-18}{\left(x-4\right)\left(x-3\right)}\)
\(=\dfrac{-3\left(x^2-5x+6\right)}{\left(x-4\right)\left(x-3\right)}=\dfrac{-3\left(x-2\right)}{x-4}=\dfrac{-3x+6}{x-4}\)
b: x^2-5x+6=0
=>x=2(nhận) hoặc x=3(loại)
Khi x=2 thì \(E=\dfrac{-3\cdot2+6}{2-4}=0\)
c: Để E là số nguyên thì -3x+12-6 chia hết cho x-4
=>\(x-4\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)
hay \(x\in\left\{5;6;2;7;1;10;-2\right\}\)
