b: \(\dfrac{A}{B}=\dfrac{-x^3+5x^2-x-8x^2+40x-8+2}{x^2-5x+1}\)

\(=-x-8+\dfrac{2}{x^2-5x+1}\)

Lời giải:
a.  

b. PT hoành độ giao điểm:

$-2x+4=3x-1$
$\Leftrightarrow x=1$

$y=3x-1=3.1-1=2$

Vậy $A(1;2)$

a) \(=\left|\sqrt{3}-3\right|+\sqrt{\left(\sqrt{3}-1\right)^2}=3-\sqrt{3}+\sqrt{3}-1=2\)

b) \(=\dfrac{\sqrt{5}+2}{5-4}-\dfrac{\sqrt{5}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=\sqrt{5}+2-\sqrt{5}=2\)

Bài 6: 

PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

              a_____2a______a_____a      (mol)

            \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

               b______3b______b_____\(\dfrac{3}{2}\)b    (mol)

Ta lập được HPT: \(\left\{{}\begin{matrix}56a+27b=22,2\\a+\dfrac{3}{2}b=\dfrac{13,44}{22,4}=0,6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,3\\b=0,2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,3\cdot56}{22,2}\cdot100\%\approx75,68\%\\\%m_{Al}=24,32\%\end{matrix}\right.\)