Áp dụng \(1+2+...+k=\frac{k\left(k+1\right)}{2}\) thì ta được :

\(\sqrt{\left[1+2+3+...+\left(n-1\right)+n\right]+\left[n+\left(n-1\right)+...+3+2+1\right]-n}=2010\)

\(\Leftrightarrow\sqrt{2.\frac{n\left(n+1\right)}{2}-n}=2010\)

\(\Leftrightarrow\sqrt{n^2}=2010\Leftrightarrow n=2010\)

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

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

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

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

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

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

ngu giữ vậy

Chó khang không trả lời thì thôi sao chửi tao ngu

Vì \(n\in Z^+\)nên\(n\left(n+1\right)\left(n+2\right)>n^3\Rightarrow\sqrt[3]{n\left(n+1\right)\left(n+2\right)}>n\)

\(\Rightarrow\sqrt[3]{n\left(n+1\right)\left(n+2\right)+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}+...+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}}>n\)(1)

Lại có:\(n^2+2n+1>n^2+2n\Rightarrow\left(n+1\right)^2>n\left(n+2\right)\Rightarrow\left(n+1\right)^3>n\left(n+1\right)\left(n+2\right)\)

\(\Rightarrow n+1>\sqrt[3]{n\left(n+1\right)\left(n+2\right)}\\ \Rightarrow\sqrt[3]{n^3+3n^2+3n+1}>\sqrt[3]{n^3+3n^2+2n}\)

\(\Rightarrow\sqrt[3]{n^3+3n^2+2n+n+1}>\sqrt[3]{n^3+3n^2+2n+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}}\)

\(\Rightarrow\sqrt[3]{\left(n+1\right)^3}>\sqrt[3]{n\left(n+1\right)\left(n+2\right)+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}}\)

Tương tự \(\Rightarrow n+1>\sqrt[3]{n\left(n+1\right)\left(n+2\right)+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}+...+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}}\)(2)

Từ (1) và (2) suy ra:

\(n< \sqrt[3]{n\left(n+1\right)\left(n+2\right)+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}+...+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}}< n+1\)

\(n\in Z^+\)nên n² < n² + 2n < n² + 2n + 1 <=> n² < n(n + 2) < (n + 1)² => n³ < n(n + 1)(n + 2) < (n + 1)³ 

=>\(n< \sqrt[3]{n\left(n+1\right)\left(n+2\right)}< n+1\)

=>\(n< \sqrt[3]{n\left(n+1\right)\left(n+2\right)}< \sqrt[3]{n\left(n+1\right)\left(n+2\right)+n}\)\(< \sqrt[3]{n\left(n+1\right)\left(n+2\right)+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}}< \sqrt[3]{n\left(n+1\right)\left(n+2\right)+n+1}\)\(=\sqrt[3]{\left(n+1\right)\left(n^2+2n+1\right)}=\sqrt[3]{\left(n+1\right)\left(n+1\right)^2}=n+1\)

=>\(n< \sqrt[3]{n\left(n+1\right)\left(n+2\right)+n}\)

\(< \sqrt[3]{n\left(n+1\right)\left(n+2\right)+\sqrt[3]{n\left(n+1\right)\left(n+2\right)+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}}}< n+1\)

Tiếp tục như vậy,ta có đpcm.

Sorry ! n² < n(n + 2) nên n³ < n(n + 1)(n + 2) (vì n < n + 1)

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

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

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

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

\(c=\lim\dfrac{\left(\dfrac{2}{\sqrt{n}}+\dfrac{1}{n}\right)\left(\dfrac{1}{\sqrt{n}}+\dfrac{3}{n}\right)}{\left(1+\dfrac{1}{n}\right)\left(1+\dfrac{2}{n}\right)}=\dfrac{0.0}{1.1}=0\)

\(d=\lim\dfrac{4-3\left(\dfrac{2}{4}\right)^n}{9.\left(\dfrac{3}{4}\right)^n+\left(\dfrac{2}{4}\right)^n}=\dfrac{4}{0}=+\infty\)

\(e=\lim\dfrac{7-25\left(\dfrac{5}{7}\right)^n+3.\left(\dfrac{1}{7}\right)^n}{12.\left(\dfrac{6}{7}\right)^n-\left(\dfrac{3}{7}\right)^n+3\left(\dfrac{1}{7}\right)^n}=\dfrac{7}{0}=+\infty\)

\(f=\lim\dfrac{n^4-4n^6}{n\left(\sqrt{n^4+1}+\sqrt{4n^6+1}\right)}=\lim\dfrac{\dfrac{1}{n^2}-6}{\sqrt{\dfrac{1}{n^6}+\dfrac{1}{n^{10}}}+\sqrt{\dfrac{4}{n^4}+\dfrac{1}{n^{10}}}}=\dfrac{-6}{0}=-\infty\)

1: \(\lim\limits_{n\rightarrow\infty}\left(\sqrt[3]{n^3+n^2+n+1}-n\right)\)

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

\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^2+n+1}{n^2\cdot\sqrt[3]{\left(1+\dfrac{1}{n}+\dfrac{1}{n^2}+\dfrac{1}{n^3}\right)^2}+n^2\cdot\sqrt[3]{1+\dfrac{1}{n}+\dfrac{1}{n^2}+\dfrac{1}{n^3}}+n^2}\)

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

\(=\dfrac{1}{1+1+1}=\dfrac{1}{3}\)

2: \(\lim\limits_{n\rightarrow\infty}\left(\sqrt{n^2+n}-\sqrt{n^2-n+1}\right)\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^2+n-n^2+n-1}{\sqrt{n^2+n}+\sqrt{n^2-n+1}}\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{2n-1}{\sqrt{n^2+n}+\sqrt{n^2-n+1}}\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{2-\dfrac{1}{n}}{\sqrt{1+\dfrac{1}{n}}+\sqrt{1-\dfrac{1}{n}+\dfrac{1}{n^2}}}\)

\(=\dfrac{2}{1+1}=\dfrac{2}{2}=1\)

\(a=\lim\dfrac{1}{\sqrt{4n+1}+2\sqrt{n}}=\dfrac{1}{\infty}=0\)

\(b=\lim n\left(\sqrt{1+\dfrac{2}{n}}-\sqrt{1-\dfrac{2}{n}}-1\right)=+\infty.\left(-1\right)=-\infty\)

\(c=\lim4^n\left(\sqrt{\left(\dfrac{9}{16}\right)^n-\left(\dfrac{3}{16}\right)^n}-1\right)=+\infty.\left(-1\right)=-\infty\)

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