\(A=\dfrac{2x+1}{\left(x-4\right)\left(x-3\right)}-\dfrac{x+3}{x-4}+\dfrac{2x+1}{x-3}\)
\(=\dfrac{2x+1}{\left(x-4\right)\left(x-3\right)}-\dfrac{\left(x+3\right)\left(x-3\right)}{\left(x-4\right)\left(x-3\right)}+\dfrac{\left(2x+1\right)\left(x-4\right)}{\left(x-4\right)\left(x-3\right)}\)
\(=\dfrac{2x+1-\left(x^2-9\right)+\left(2x^2-8x+x-4\right)}{\left(x-4\right)\left(x-3\right)}\)
\(=\dfrac{2x+1-x^2+9+2x^2-7x-4}{\left(x-4\right)\left(x-3\right)}\)
\(=\dfrac{x^2-5x+6}{\left(x-4\right)\left(x-3\right)}\)
\(=\dfrac{\left(x^2-2x\right)-\left(3x-6\right)}{\left(x-4\right)\left(x-3\right)}\)
\(=\dfrac{\left(x-3\right)\left(x-2\right)}{\left(x-4\right)\left(x-3\right)}=\dfrac{x-2}{x-4}\)
Tiếp tục câu 2
b.
$x^2+20=9x$
$\Leftrightarrow x^2-9x+20=0$
$\Leftrightarrow (x-4)(x-5)=0$
$\Rightarrow x=5$ (do $x\neq 4$)
Khi đó: $A=\frac{x-2}{x-4}=\frac{5-2}{5-4}=3$
c.
$B=A(x^2-5x+4)=\frac{x-2}{x-4}.(x-1)(x-4)=(x-1)(x-2)$
$=x^2-3x+2=(x-1,5)^2-0,25\geq -0,25$
Vậy $B_{\min}=-0,25$ khi $x=1,5$
