1b..
$\mathop {\lim }\limits_{x \to 3}\left(\dfrac{4x-3}{2x+7}\right)^5=\left(\dfrac{4.3-3}{2.3+7}\right)^5=\dfrac{9^5}{13^5}$

5.
     $\mathop {\lim }\limits_{x \to -\infty}\dfrac{1}{2x^3-x^2+3x+5}$
$=\mathop {\lim }\limits_{x \to -\infty}\dfrac{\dfrac{1}{x^3}}{2-\dfrac{1}{x}+\dfrac{3}{x^2}+\dfrac{5}{x^3}}=0$

4.
     $\mathop {\lim }\limits_{x \to -\infty}\dfrac{x^2-5x+2}{2|x|+1}$
$=\mathop {\lim }\limits_{x \to -\infty}\dfrac{|x|^2+5|x|+2}{2|x|+1}$
$=\mathop {\lim }\limits_{x \to -\infty}\dfrac{|x|+5+\dfrac{2}{|x|}}{2+\dfrac{1}{|x|}}=+\infty$

3.
     $\mathop {\lim }\limits_{x \to -\infty}\dfrac{2x^4-x-1}{x^2+x+1}$
$=\mathop {\lim }\limits_{x \to -\infty}\dfrac{2x^2-\dfrac{1}{x}-\dfrac{1}{x^2}}{1+\dfrac{1}{x}+\dfrac{1}{x^2}}=+\infty$

2.
     $\mathop {\lim }\limits_{x \to -\infty}\dfrac{\sqrt{x^4-x}}{1-2x}$
$=\mathop {\lim }\limits_{x \to -\infty}\dfrac{\sqrt{x^2-\dfrac{1}{x}}}{-\dfrac{1}{x}+2}=+\infty$

1.
     $\mathop {\lim }\limits_{x \to +\infty}\dfrac{x^3-5}{x^2+1}$
$=\mathop {\lim }\limits_{x \to +\infty}\dfrac{x-\dfrac{5}{x^2}}{1+\dfrac{1}{x^2}}$
$=\mathop {\lim }\limits_{x \to +\infty}\dfrac{x(1-\dfrac{5}{x^3})}{1+\dfrac{1}{x^2}}=+\infty$

Điều kiện: $\left\{\begin{array}{l}-xy-2x+y+2>0,x^2-2x+1>0,y+5>0,x+4>0\\0<1-x\ne1,0<2+y\ne1\end{array}\right.$
Hệ đã cho tương đương với:
      $\left\{\begin{array}{l}2\log_{1-x}[(1-x)(y+2)]+2\log_{2+y}(1-x)=6\\\log_{1-x}(y+5)-\log_{2+y}(x+4)=1\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}\log_{1-x}(y+2)+\log_{2+y}(1-x)=2&(1)\\\log_{1-x}(y+5)-\log_{2+y}(x+4)=1&(2)\end{array}\right.$
Đặt: $\log_{2+y}(1-x)=t$ thì $(1)$ trở thành:
         $t+\dfrac{1}{t}=2 \Leftrightarrow (t-1)^2=0 \Leftrightarrow t=1$
Với $t=1$ ta có: $1-x=y+2 \Leftrightarrow y=-x-1$
Thế vào $(2)$ ta được:
$\log_{1-x}(-x+4)-\log_{1-x}(x+4)=1 \Leftrightarrow \log_{1-x}\dfrac{-x+4}{x+4}=1 \Leftrightarrow \dfrac{-x+4}{x+4}=1 \Leftrightarrow \left[\begin{array}{l}x=0\\x=-2\end{array}\right.$
Với $x=0 \Rightarrow y=-1$, loại.
Với $x=-2 \Rightarrow y=1$, thỏa mãn.
Vậy: $(x;y)=(-2;1)$

6.
     $\mathop {\lim }\limits_{x \to 1} \dfrac{\sqrt{2x-1}+x^2-3x+1}{\sqrt[3]{x-2}+x^2-x+1}$
$=\mathop {\lim }\limits_{x \to 1} \dfrac{(\sqrt{2x-1}-1)+x^2-3x+2}{(1-\sqrt[3]{2-x})+x^2-x}$
$=\mathop {\lim }\limits_{x \to 1} \dfrac{\dfrac{2x-2}{\sqrt{2x-1}+1}+(x-1)(x-2)}{\dfrac{x-1}{1+\sqrt[3]{2-x}+\sqrt[3]{(2-x)^2}}+x(x-1)}$
$=\mathop {\lim }\limits_{x \to 1} \dfrac{\dfrac{2}{\sqrt{2x-1}+1}+x-2}{\dfrac{1}{1+\sqrt[3]{2-x}+\sqrt[3]{(2-x)^2}}+x}=0$