\(\lim\left(3n-\sqrt{9n^2+1}\right)=\lim\dfrac{-1}{3n+\sqrt{9n^2+1}}=\lim\dfrac{-\dfrac{1}{n}}{3+\sqrt{9+\dfrac{1}{n^2}}}=\dfrac{0}{3+3}=0\)

\(\lim\left(\sqrt[3]{n^3-2n^2}-n\right)=\lim\dfrac{-2n^2}{\sqrt[3]{\left(n^3-2n^2\right)^2}+n\sqrt[3]{n^3-2n^2}+n^2}\)

\(=\lim\dfrac{-2}{\sqrt[3]{\left(1-\dfrac{2}{n}\right)^2}+\sqrt[3]{1-\dfrac{2}{n}}+1}=\dfrac{-2}{1+1+1}=-\dfrac{2}{3}\)

a) lim \(\frac{\left(2n^2-3n+5\right)\left(2n+1\right)}{\left(4-3n\right)\left(2n^2+n+1\right)}\)

= lim \(\frac{\left(2-\frac{3}{n}+\frac{5}{n^2}\right)\left(2+\frac{1}{n}\right)}{\left(\frac{4}{n}-3\right)\left(2+\frac{1}{n}+\frac{1}{n^2}\right)}=\frac{4}{-6}=-\frac{2}{3}\)

b)lim ( \(\frac{\sqrt{n^4+1}}{n}-\frac{\sqrt{4n^6+2}}{n^2}\))

= lim ( \(\frac{n\sqrt{n^4+1}-\sqrt{4n^6+2}}{n^2}\) )

= lim \(\frac{\left(n^6+n^2\right)-\left(4n^6+2\right)}{n^2\left(n\sqrt{n^4+1}+\sqrt{4n^2+2}\right)}\)

= lim \(\frac{-3n^6+n^2+2}{n^3\sqrt{n^4+1}+n^2\sqrt{4n^2+2}}\)

= lim \(\frac{-3n\left(1-\frac{1}{n^4}-\frac{2}{n^6}\right)}{\sqrt{1+\frac{1}{n^4}}+\frac{1}{n^2}\sqrt{4+\frac{2}{n^2}}}\)

= lim \(-3n=-\infty\)

c) lim \(\frac{2n+3}{\sqrt{9n^2+3}-\sqrt[3]{2n^2-8n^3}}\)

= lim\(\frac{2+\frac{3}{n}}{\sqrt{9+\frac{3}{n^2}}-\sqrt[3]{\frac{2}{n}-8}}=\frac{2}{3+2}=\frac{2}{5}\)

Mình sửa lại chút nhé. tìm x,  y là các số hữu tỉ

\(a=\lim4^n\left(1-\left(\dfrac{3}{4}\right)^n\right)=+\infty.1=+\infty\)

\(b=\lim\left(4^n+2.2^n+1-4^n\right)=\lim2^n\left(2+\dfrac{1}{2^n}\right)=+\infty.2=+\infty\)

\(c=limn^3\left(\sqrt{\dfrac{2}{n}-\dfrac{3}{n^4}+\dfrac{11}{n^6}}-1\right)=+\infty.\left(-1\right)=-\infty\)

\(d=\lim n\left(\sqrt{2+\dfrac{1}{n^2}}-\sqrt{3-\dfrac{1}{n^2}}\right)=+\infty\left(\sqrt{2}-\sqrt{3}\right)=-\infty\)

\(e=\lim\dfrac{3n\sqrt{n}+1}{\sqrt{n^2+3n\sqrt{n}+1}+n}=\lim\dfrac{3\sqrt{n}+\dfrac{1}{n}}{\sqrt{1+\dfrac{3}{\sqrt{n}}+\dfrac{1}{n^2}}+1}=\dfrac{+\infty}{2}=+\infty\)

\(=\lim\dfrac{\left(2+\dfrac{1}{n\sqrt{n}}\right)\left(1+\dfrac{3}{\sqrt{n}}\right)}{\left(1+\dfrac{1}{n}\right)\left(1+\dfrac{2}{n}\right)}=\dfrac{2.1}{1.1}=2\)

\(\Leftrightarrow\lim\limits\dfrac{9n^2-9n^2+\left(a+1\right)n}{3n+\sqrt{9n^2-\left(a+1\right)n}}=6\)

\(\Leftrightarrow\lim\limits\dfrac{\dfrac{n\left(a+1\right)}{n}}{\dfrac{3n}{n}+\sqrt{\dfrac{9n^2}{n^2}-\dfrac{\left(a+1\right)n}{n^2}}}=6\)

\(\Leftrightarrow\dfrac{a+1}{3+3}=6\Leftrightarrow a=35\)