b.
Ta có:
      $3a+4b\,\vdots\,11$
$\Rightarrow 4(3a+4b)\,\vdots\,11$
$\Rightarrow 12a+16b\,\vdots\,11$
$\Rightarrow 12a+16b-11a-11b\,\vdots\,11$
$\Rightarrow a+5b\,\vdots\,11$

a.
Ta có:
      $100a+b\,\vdots\,7$
$\Rightarrow 4(100a+b)\,\vdots\,7$
$\Rightarrow 400a+4b\,\vdots\,7$
$\Rightarrow 400a+4b-57.7a\,\vdots\,7$
$\Rightarrow a+4b\,\vdots\,7$

6.
     $\mathop {\lim }\limits_{x \to 1}\dfrac{\sqrt{4x+5}+\sqrt{3x+1}-5}{x-1}$
$=\mathop {\lim }\limits_{x \to 1}\dfrac{(\sqrt{4x+5}-3)+(\sqrt{3x+1}-2)}{x-1}$
$=\mathop {\lim }\limits_{x \to 1}\dfrac{\dfrac{4x-4}{\sqrt{4x+5}+3}+\dfrac{3x-3}{\sqrt{3x+1}+2}}{x-1}$
$=\mathop {\lim }\limits_{x \to 1}\left(\dfrac{4}{\sqrt{4x+5}+3}+\dfrac{3}{\sqrt{3x+1}+2}\right)=\dfrac{17}{12}$

5.
     $\mathop {\lim }\limits_{x \to 2}\dfrac{\sqrt[3]{3x+2}-\sqrt{x+2}}{x^2-x-2}$
$=\mathop {\lim }\limits_{x \to 2}\dfrac{(\sqrt[3]{3x+2}-2)-(\sqrt{x+2}-2)}{x^2-x-2}$
$=\mathop {\lim }\limits_{x \to 2}\dfrac{\dfrac{3x-6}{\sqrt[3]{(3x+2)^2}+2\sqrt[3]{3x+2}+4}-\dfrac{x-2}{\sqrt{x+2}+2}}{(x-2)(x+1)}$
$=\mathop {\lim }\limits_{x \to 2}\dfrac{\dfrac{3}{\sqrt[3]{(3x+2)^2}+2\sqrt[3]{3x+2}+4}-\dfrac{1}{\sqrt{x+2}+2}}{x+1}=0$

4.
     $\mathop {\lim }\limits_{x \to 1}\dfrac{2-\sqrt{2x-1}\sqrt[3]{5x+3}}{x-1}$
$=\mathop {\lim }\limits_{x \to 1}\dfrac{(2-2\sqrt{2x-1})+\sqrt{2x-1}(2-\sqrt[3]{5x+3})}{x-1}$
$=\mathop {\lim }\limits_{x \to 1}\dfrac{\dfrac{2(2-2x)}{1+\sqrt{2x-1}}+\dfrac{\sqrt{2x-1}(5-5x)}{4+2\sqrt[3]{5x+3}+\sqrt[3]{(5x+3)^2}}}{x-1}$
$=\mathop {\lim }\limits_{x \to 1}\left(\dfrac{-4}{1+\sqrt{2x-1}}-\dfrac{5\sqrt{2x-1}}{4+2\sqrt[3]{5x+3}+\sqrt[3]{(5x+3)^2}}\right)=\dfrac{-29}{12}$

3.
     $\mathop {\lim }\limits_{x \to 0}\dfrac{\sqrt[m]{1+ax}\sqrt[n]{1+bx}-1}{x}$
$=\mathop {\lim }\limits_{x \to 0}\dfrac{\sqrt[m]{1+ax}(\sqrt[n]{1+bx}-1)+(\sqrt[m]{1+ax}-1)}{x}$
$=\mathop {\lim }\limits_{x \to 0}\dfrac{\dfrac{bx\sqrt[m]{1+ax}}{\sqrt[n]{(1+bx)^{n-1}}+\ldots+\sqrt[n]{1+bx}+1}+\dfrac{ax}{\sqrt[m]{(1+ax)^{m-1}}+\ldots+\sqrt[m]{1+ax}+1}}{x}$
$=\mathop {\lim }\limits_{x \to 0}\left(\dfrac{b\sqrt[m]{1+ax}}{\sqrt[n]{(1+bx)^{n-1}}+\ldots+\sqrt[n]{1+bx}+1}+\dfrac{a}{\sqrt[m]{(1+ax)^{m-1}}+\ldots+\sqrt[m]{1+ax}+1}\right)=\dfrac{a}{m}+\dfrac{b}{n}$

2.
     $\mathop {\lim }\limits_{x \to 0}\dfrac{\sqrt{1+2x}\sqrt[3]{1+3x}\sqrt[3]{1+4x}-1}{x}$
$=\mathop {\lim }\limits_{x \to 0}\dfrac{\sqrt{1+2x}\sqrt[3]{1+3x}(\sqrt[3]{1+4x}-1)+\sqrt{1+2x}(\sqrt[3]{1+3x}-1)+(\sqrt{1+2x}-1)}{x}$
$=\mathop {\lim }\limits_{x \to 0}\dfrac{\dfrac{4x\sqrt{1+2x}\sqrt[3]{1+3x}}{\sqrt[3]{(1+4x)^2}+\sqrt[3]{1+4x}+1}+\dfrac{3x\sqrt{1+2x}}{\sqrt[3]{(1+3x)^2}+\sqrt[3]{1+3x}+1}+\dfrac{2x}{\sqrt{1+2x}+1}}{x}$
$=\mathop {\lim }\limits_{x \to 0}\left(\dfrac{4\sqrt{1+2x}\sqrt[3]{1+3x}}{\sqrt[3]{(1+4x)^2}+\sqrt[3]{1+4x}+1}+\dfrac{3\sqrt{1+2x}}{\sqrt[3]{(1+3x)^2}+\sqrt[3]{1+3x}+1}+\dfrac{2}{\sqrt{1+2x}+1}\right)=\dfrac{10}{3}$