Đặt \(a=\sqrt3-\sqrt2\)

=>\(a^2=\left(\sqrt3-\sqrt2\right)^2=5-2\sqrt6\)

=>\(-a^2=2\sqrt6-5\)

\(3a=3\left(\sqrt3-\sqrt2\right)=3\sqrt3-3\sqrt2\)

\(\frac{1}{a}=\frac{1}{\sqrt3-\sqrt2}=\frac{\sqrt3+\sqrt2}{\left(\sqrt3-\sqrt2\right)\left(\sqrt3+\sqrt2\right)}=\sqrt3+\sqrt2\)

Khi đó, ta sẽ có: \(u_{n+1} = a u_n^2 - a^2 u_n + 3a\) (1)

Đặt \(v_n = u_n + \sqrt{2}\)

=>\(u_{n}=v_{n}-\sqrt{2}\)

Thay vào (1), ta có: \(v_{n+1} - \sqrt{2} = a(v_n - \sqrt{2})^2 - a^2(v_n - \sqrt{2}) + 3a\)

=>\(v_{n+1} = a v_n^2 - v_n + \frac{1}{a}\) , với số hạng đầu tiên là \(v_1 = u_1 + \sqrt{2} = \sqrt{3} + 2\sqrt{2}\)

Đặt \(x_n = v_n - \frac{1}{a}\)

\(x_1 = v_1 - \frac{1}{a} = (\sqrt{3} + 2\sqrt{2}) - (\sqrt{3} + \sqrt{2}) = \sqrt{2}\)

\(x_{n+1}=v_{n+1}-\frac{1}{a}\)

\(=av_{n}^2-v_{n}\)

\(=v_{n}(av_{n}-1)\)

\(=\left(x_{n}+\frac{1}{a}\right)ax_{n}\)

\(=x_{n}+ax_{n}^2\)

\(v_n = x_n + \frac{1}{a}\)

\(\frac{1}{x_n}-\frac{1}{x_{n+1}}=\frac{x_{n+1} - x_n}{x_n x_{n+1}}\)

\(=\frac{a x_n^2}{x_n(x_n + a x_n^2)}\)

\(=\frac{a x_n}{1 + a x_n}=\frac{1}{x_n + \frac{1}{a}}=\frac{1}{v_n}\)

=>\(\frac{1}{v_n} = \frac{1}{x_n} - \frac{1}{x_{n+1}}\)

\(S_n = \sum_{i=1}^n \frac{1}{u_i + \sqrt{2}} = \sum_{i=1}^n \frac{1}{v_i}\)

\(=\left(\frac{1}{x_1}-\frac{1}{x_2}\right)+\left(\frac{1}{x_2}-\frac{1}{x_3}\right)+\ldots+\left(\frac{1}{x_n}-\frac{1}{x_{n+1}}\right)=\frac{1}{x_1}-\frac{1}{x_{n+1}}\)

Ta có: \(\begin{cases}x_1=\sqrt2>0\\ a=\sqrt3-\sqrt2>0\end{cases}\)

=>(\(x_{n}\) ) là dãy tăng và \(\lim_{n\to\infty}x_{n}=+\infty\)

=>\(\lim_{n \to \infty} S_n = \lim_{n \to \infty} \left(\frac{1}{x_1} - \frac{1}{x_{n+1}}\right) = \frac{1}{x_1} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}\)

=>\(\lim \left(\sum_{i=1}^n \frac{1}{u_i + \sqrt{2}}\right) = \frac{\sqrt{2}}{2}\)

\(=\lim\limits\dfrac{n^2+an+5-n^2-1}{\sqrt{n^2+an+5}+\sqrt{n^2+1}}=\lim\limits\dfrac{an+4}{\sqrt{n^2+an+5}+\sqrt{n^2+1}}\)

\(=\lim\limits\dfrac{\dfrac{an}{n}+\dfrac{4}{n}}{\sqrt{\dfrac{n^2}{n^2}+\dfrac{an}{n^2}+\dfrac{5}{n^2}}+\sqrt{\dfrac{n^2}{n^2}+\dfrac{1}{n^2}}}=\dfrac{a}{1+1}=\dfrac{a}{2}\)

\(\lim\limits\left(u_n\right)=-1\Rightarrow\dfrac{a}{2}=-1\Rightarrow a=-2\)

\(\lim\limits\left(2-3n\right)^4\left(n+1\right)^3=\lim n^7\left(3-\dfrac{2}{n}\right)^4\left(1+\dfrac{1}{n}\right)^3=+\infty\)

\(\lim\left(\sqrt[3]{n+4}-\sqrt[3]{n+1}\right)=\lim\dfrac{3}{\sqrt[3]{\left(n+4\right)^2}+\sqrt[3]{\left(n+4\right)\left(n+1\right)}+\sqrt[3]{\left(n+1\right)^2}}=0\)

\(\lim\left(\sqrt[3]{8n^3+3n^2+4}-2n+6\right)=\lim\dfrac{8n^3+3n^2+4-\left(2n-6\right)^3}{\sqrt[3]{\left(8n^3+3n^2+4\right)^2}+\left(2n-6\right)\sqrt[3]{8n^3+3n^2+4}+\left(2n-6\right)^2}\)

\(=\lim\dfrac{75n^2-216n+220}{\sqrt[3]{\left(8n^3+3n^2+4\right)^2}+\left(2n-6\right)\sqrt[3]{8n^3+3n^2+4}+\left(2n-6\right)^2}\)

\(=\lim\dfrac{75-\dfrac{216}{n}+\dfrac{220}{n^2}}{\sqrt[3]{\left(8+\dfrac{3}{n}+\dfrac{4}{n^3}\right)^2}+\left(2-\dfrac{6}{n}\right)\sqrt[3]{8+\dfrac{3}{n}+\dfrac{4}{n^3}}+\left(2-\dfrac{6}{n}\right)^2}\)

\(=\dfrac{75}{\sqrt[3]{8^2}+2.\sqrt[3]{8}+2^2}=...\)

d.

\(\lim\left(\sqrt[3]{8n^3+3n^2-2}+\sqrt[3]{5n^2-8n^3}\right)\)

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

\(=\lim\dfrac{8n^3+3n^2-2-\left(8n^3-5n^2\right)}{\sqrt[3]{\left(8n^3+3n^2-2\right)^2}+\sqrt[3]{\left(8n^3+3n^2-2\right)\left(8n^3-5n^2\right)}+\sqrt[3]{8n^3-5n^2}}\)

\(=\lim\dfrac{8n^2-2}{\sqrt[3]{\left(8n^3+3n^2-2\right)^2}+\sqrt[3]{\left(8n^3+3n^2-2\right)\left(8n^3-5n^2\right)}+\sqrt[3]{8n^3-5n^2}}\)

\(=lim\dfrac{8-\dfrac{2}{n^2}}{\sqrt[3]{\left(8+\dfrac{3}{n}-\dfrac{2}{n^3}\right)^2}+\sqrt[3]{\left(8+\dfrac{3}{n}-\dfrac{2}{n^3}\right)\left(8-\dfrac{5}{n}\right)}+\sqrt[3]{\left(8-\dfrac{5}{n}\right)^2}}\)

\(=\dfrac{8}{\sqrt[3]{8^2}+\sqrt[3]{8.8}+\sqrt[3]{8^2}}=...\)

Hiện tại mới nghĩ được câu b thôi

b/ \(u_1=\dfrac{1}{2};u_2=\dfrac{1}{2-\dfrac{1}{2}}=\dfrac{2}{3};u_3=\dfrac{1}{2-\dfrac{2}{3}}=\dfrac{3}{4}...\)

Nhận thấy \(u_n=\dfrac{n}{n+1}\) , ta sẽ chứng minh bằng phương pháp quy nạp

\(n=k\Rightarrow u_k=\dfrac{k}{k+1}\)

Chứng minh cũng đúng với \(\forall n=k+1\)

\(\Rightarrow u_{k+1}=\dfrac{k+1}{k+2}\)

Ta có: \(u_{k+1}=\dfrac{1}{2-u_k}=\dfrac{1}{2-\dfrac{k}{k+1}}=\dfrac{k+1}{k+2}\)

Vậy biểu thức đúng với \(\forall n\in N\left(n\ne0\right)\)

\(\Rightarrow limu_n=lim\dfrac{n}{n+1}=lim\dfrac{1}{1+\dfrac{1}{n}}=1\)

\(a=\lim\dfrac{5n\left(n+\sqrt{n^2-n-1}\right)}{n+1}=\lim\dfrac{5\left(n+\sqrt{n^2-n-1}\right)}{1+\dfrac{1}{n}}=\dfrac{+\infty}{1}=+\infty\)

\(b=\lim\dfrac{\sqrt{\dfrac{1}{n}+\sqrt{\dfrac{1}{n^3}+\dfrac{1}{n^4}}}}{1-\dfrac{1}{\sqrt{n}}}=\dfrac{0}{1}=0\)

\(c=\lim\dfrac{\sqrt{2n^2-1+\dfrac{7}{n^2}}}{3+\dfrac{5}{n}}=\dfrac{+\infty}{3}=+\infty\)

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

\(u_2=\sqrt{2}\left(2+3\right)-3=5\sqrt{2}-3\)

\(u_3=\sqrt{\dfrac{3}{2}}.5\sqrt{2}-3=5\sqrt{3}-3\)

\(u_4=\sqrt{\dfrac{4}{3}}.5\sqrt{3}-3=5\sqrt{4}-3\)

....

\(\Rightarrow u_n=5\sqrt{n}-3\)

\(\Rightarrow\lim\limits\dfrac{u_n}{\sqrt{n}}=\lim\limits\dfrac{5\sqrt{n}-3}{\sqrt{n}}=5\)

\(\lim\limits_{x\rightarrow0}\dfrac{3x^2+2-\left(2-2x\right)}{x\left(\sqrt{3x^2+2}+\sqrt{2-2x}\right)}=\lim\limits_{x\rightarrow0}\dfrac{x\left(3x+2\right)}{x\left(\sqrt{3x^2+2}+\sqrt{2-2x}\right)}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{3x+2}{\sqrt{3x^2+2}+\sqrt{2-2x}}=\dfrac{2}{2\sqrt{2}}=\dfrac{\sqrt{2}}{2}\)

\(\Rightarrow\left\{{}\begin{matrix}a=1\\b=2\end{matrix}\right.\)