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\(u_2=\sqrt{2}\left(2+3\right)-3=5\sqrt{2}-3\)
\(u_3=\sqrt{\dfrac{3}{2}}.5\sqrt{2}-3=5\sqrt{3}-3\)
\(u_4=\sqrt{\dfrac{4}{3}}.5\sqrt{3}-3=5\sqrt{4}-3\)
....
\(\Rightarrow u_n=5\sqrt{n}-3\)
\(\Rightarrow\lim\limits\dfrac{u_n}{\sqrt{n}}=\lim\limits\dfrac{5\sqrt{n}-3}{\sqrt{n}}=5\)
\(\lim\limits\left(2-3n\right)^4\left(n+1\right)^3=\lim n^7\left(3-\dfrac{2}{n}\right)^4\left(1+\dfrac{1}{n}\right)^3=+\infty\)
\(\lim\left(\sqrt[3]{n+4}-\sqrt[3]{n+1}\right)=\lim\dfrac{3}{\sqrt[3]{\left(n+4\right)^2}+\sqrt[3]{\left(n+4\right)\left(n+1\right)}+\sqrt[3]{\left(n+1\right)^2}}=0\)
\(\lim\left(\sqrt[3]{8n^3+3n^2+4}-2n+6\right)=\lim\dfrac{8n^3+3n^2+4-\left(2n-6\right)^3}{\sqrt[3]{\left(8n^3+3n^2+4\right)^2}+\left(2n-6\right)\sqrt[3]{8n^3+3n^2+4}+\left(2n-6\right)^2}\)
\(=\lim\dfrac{75n^2-216n+220}{\sqrt[3]{\left(8n^3+3n^2+4\right)^2}+\left(2n-6\right)\sqrt[3]{8n^3+3n^2+4}+\left(2n-6\right)^2}\)
\(=\lim\dfrac{75-\dfrac{216}{n}+\dfrac{220}{n^2}}{\sqrt[3]{\left(8+\dfrac{3}{n}+\dfrac{4}{n^3}\right)^2}+\left(2-\dfrac{6}{n}\right)\sqrt[3]{8+\dfrac{3}{n}+\dfrac{4}{n^3}}+\left(2-\dfrac{6}{n}\right)^2}\)
\(=\dfrac{75}{\sqrt[3]{8^2}+2.\sqrt[3]{8}+2^2}=...\)
d.
\(\lim\left(\sqrt[3]{8n^3+3n^2-2}+\sqrt[3]{5n^2-8n^3}\right)\)
\(=\lim\left(\sqrt[3]{8n^3+3n^2-2}-\sqrt[3]{8n^3-5n^2}\right)\)
\(=\lim\dfrac{8n^3+3n^2-2-\left(8n^3-5n^2\right)}{\sqrt[3]{\left(8n^3+3n^2-2\right)^2}+\sqrt[3]{\left(8n^3+3n^2-2\right)\left(8n^3-5n^2\right)}+\sqrt[3]{8n^3-5n^2}}\)
\(=\lim\dfrac{8n^2-2}{\sqrt[3]{\left(8n^3+3n^2-2\right)^2}+\sqrt[3]{\left(8n^3+3n^2-2\right)\left(8n^3-5n^2\right)}+\sqrt[3]{8n^3-5n^2}}\)
\(=lim\dfrac{8-\dfrac{2}{n^2}}{\sqrt[3]{\left(8+\dfrac{3}{n}-\dfrac{2}{n^3}\right)^2}+\sqrt[3]{\left(8+\dfrac{3}{n}-\dfrac{2}{n^3}\right)\left(8-\dfrac{5}{n}\right)}+\sqrt[3]{\left(8-\dfrac{5}{n}\right)^2}}\)
\(=\dfrac{8}{\sqrt[3]{8^2}+\sqrt[3]{8.8}+\sqrt[3]{8^2}}=...\)
Một câu thôi: Liên hợp
\(\dfrac{1}{2\sqrt{1}+\sqrt{2}}=\dfrac{2.1-\sqrt{2}}{2^2-2}=\dfrac{2-\sqrt{2}}{2}=1-\dfrac{1}{\sqrt{2}}\)
\(\dfrac{1}{3\sqrt{2}+2\sqrt{3}}=\dfrac{3\sqrt{2}-2\sqrt{3}}{9.2-4.3}=\dfrac{3\sqrt{2}-2\sqrt{3}}{6}=\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}\)
\(\Rightarrow\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
Nên chứng minh bằng quy nạp mạnh cho chặt chẽ, giờ tui buồn ngủ quá nên bạn tự chứng minh nha :(
\(\Rightarrow u_n=1-\dfrac{1}{\sqrt{n+1}}=\dfrac{\sqrt{n+1}-1}{\sqrt{n+1}}\Rightarrow\lim\limits\left(u_n\right)=\lim\limits\dfrac{\sqrt{\dfrac{n}{n}+\dfrac{1}{n}}-\dfrac{1}{\sqrt{n}}}{\sqrt{\dfrac{n}{n}+\dfrac{1}{n}}}=1\)
4: Ta có: \(\dfrac{6}{1-\sqrt{3}}-\dfrac{3\sqrt{3}+3}{\sqrt{3}+1}\)
\(=-3-3\sqrt{3}-3\)
\(=-6-3\sqrt{3}\)
`c)root{3}{4}.root{3}{1-sqrt3}.root{6}{(sqrt3+1)^2}`
`=root{3}{4(1-sqrt3)}.root{3}{1+sqrt3}`
`=root{3}{4(1-sqrt3)(1+sqrt3)}`
`=root{3}{4(1-3)}=-2`
`d)2/(root{3}{3}-1)-4/(root{9}-root{3}{3}+1)`
`=(2(root{3}{9}+root{3}{3}+1))/(3-1)-(4(root{3}{3}+1))/(3+1)`
`=root{3}{9}+root{3}{3}+1-root{3}{3}-1`
`=root{3}{9}`
`a)root{3}{8sqrt5-16}.root{3}{8sqrt5+16}`
`=root{3}{(8sqrt5-16)(8sqrt5+16)}`
`=root{3}{320-256}`
`=root{3}{64}=4`
`b)root{3}{7-5sqrt2}-root{6}{8}`
`=root{3}{1-3.sqrt{2}+3.2.1-2sqrt2}-root{6}{(2)^3}`
`=root{3}{(1-sqrt2)^3}-sqrt2`
`=1-sqrt2-sqrt2=1-2sqrt2`
