Chứng minh rẳng biểu thức sau luôn nhận giá trị nguyên dương với mọi n nguyên dương
\(T=\left(\sqrt{2n^2+2n+1}+\sqrt{2n^2-2n+1}\right)\sqrt{4n^2+2-2\sqrt{4n^2+1}}\)
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Những câu hỏi liên quan
10 tháng 12 2021
\(f\left(n\right)=\dfrac{2n-1+2n+1+\sqrt{\left(2n+1\right)\left(2n+1\right)}}{\sqrt{2n+1}+\sqrt{2n-1}}\\ f\left(n\right)=\dfrac{\left(\sqrt{2n+1}-\sqrt{2n-1}\right)\left(2n-1+2n+1+\sqrt{\left(2n+1\right)\left(2n+1\right)}\right)}{2n+1-2n+1}\\ f\left(n\right)=\dfrac{\left(\sqrt{2n+1}\right)^3-\left(\sqrt{2n+1}\right)^3}{2}=\dfrac{\left(2n+1\right)\sqrt{2n+1}-\left(2n-1\right)\sqrt{2n+1}}{2}\)
\(\Leftrightarrow f\left(1\right)+f\left(2\right)+...+f\left(40\right)=\dfrac{3\sqrt{3}-1\sqrt{1}+5\sqrt{5}-3\sqrt{3}+...+81\sqrt{81}-79\sqrt{79}}{2}\\ =\dfrac{81\sqrt{81}-1\sqrt{1}}{2}=\dfrac{9^3-1}{2}=364\)
NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 1 2021
\(a=\lim\left(\dfrac{2n^3\left(5n+1\right)+\left(2n^2+3\right)\left(1-5n^2\right)}{\left(2n^2+3\right)\left(5n+1\right)}\right)\)
\(=\lim\left(\dfrac{2n^3-13n^2+3}{\left(2n^2+3\right)\left(5n+1\right)}\right)=\lim\dfrac{2-\dfrac{13}{n}+\dfrac{3}{n^3}}{\left(2+\dfrac{3}{n^2}\right)\left(5+\dfrac{1}{n}\right)}=\dfrac{2}{2.5}=\dfrac{1}{5}\)
\(b=\lim\left(\dfrac{n-2}{\sqrt{n^2+n}+\sqrt{n^2+2}}\right)=\lim\dfrac{1-\dfrac{2}{n}}{\sqrt{1+\dfrac{1}{n}}+\sqrt{1+\dfrac{2}{n}}}=\dfrac{1}{2}\)
\(c=\lim\dfrac{\sqrt{1+\dfrac{3}{n^3}-\dfrac{2}{n^4}}}{2-\dfrac{2}{n}+\dfrac{3}{n^2}}=\dfrac{1}{2}\)
\(d=\lim\dfrac{\sqrt{1-\dfrac{4}{n}}-\sqrt{4+\dfrac{1}{n^2}}}{\sqrt{3+\dfrac{1}{n^2}}-1}=\dfrac{1-2}{\sqrt{3}-1}=-\dfrac{1+\sqrt{3}}{2}\)
4 tháng 10 2019
2. Ta có: P = 2x2 + y2 - 4x - 4y + 10
P = 2(x2 - 2x + 1) + (y2 - 4y + 4) + 4
P = 2(x - 1)2 + (y - 2)2 + 4 \(\ge\)4 \(\forall\)x;y
=> P luôn dương với mọi biến x;y
3 Ta có:
(2n + 1)(n2 - 3n - 1) - 2n3 + 1
= 2n3 - 6n2 - 2n + n2 - 3n - 1 - 2n3 + 1
= -5n2 - 5n = -5n(n + 1) \(⋮\)5 \(\forall\)n \(\in\)Z
NN
14 tháng 9 2020
Ta có: \(VT=\sqrt{\left(2n+1\right)^2}+\sqrt{4n^2}=\sqrt{\left(2n+1\right)^2}+\sqrt{\left(2n\right)^2}\)
\(=\left|2n+1\right|+\left|2n\right|\)
Vì \(n\inℕ\)\(\Rightarrow2n+1>0\); \(2n\ge0\)
\(\Rightarrow\left|2n+1\right|=2n+1\)và \(\left|2n\right|=2n\)
\(\Rightarrow VT=2n+1+2n=4n+1\)
Ta có: \(VP=\left(2n+1\right)^2-4n^2=\left(2n+1\right)^2-\left(2n\right)^2\)
\(=\left(2n+1-2n\right)\left(2n+1+2n\right)=4n+1\)
\(\Rightarrow VT=VP\)\(\Rightarrowđpcm\)
HT
6 tháng 2 2021
\(\lim\limits\dfrac{\sqrt{\dfrac{an^3}{n^3}+\dfrac{n^2}{n^3}+\dfrac{1}{n^3}}-\sqrt{\dfrac{2n^3}{n^3}+\dfrac{n^2}{n^3}}}{\sqrt{\dfrac{4n^3}{n^3}+\dfrac{3n}{n^3}}}=\dfrac{\sqrt{a}-\sqrt{2}}{2}\le\sqrt{2}\)
\(\Rightarrow\sqrt{a}\le2\sqrt{2}+\sqrt{2}\Rightarrow-\left(2\sqrt{2}+\sqrt{2}\right)^2\le a\le\left(2\sqrt{2}+\sqrt{2}\right)^2\)
Dung ko nhi :D?
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 2 2022
\(\lim\left(\sqrt{4n^2+5n}-2n\right)=\lim\dfrac{5n}{\sqrt{4n^2+5n}+2n}=\lim\dfrac{5}{\sqrt{4+\dfrac{5}{n}}+2}=\dfrac{5}{\sqrt{4+0}+2}=\dfrac{5}{4}\)
\(\lim\left(\sqrt{2n+1}-\sqrt{n}\right)=\lim\sqrt{n}\left(\sqrt{2+\dfrac{1}{n}}-1\right)=+\infty.\left(\sqrt{2}-1\right)=+\infty\) (do \(\sqrt{2}-1>0\))
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 2 2021
\(a=\lim\dfrac{1}{\sqrt{4n+1}+2\sqrt{n}}=\dfrac{1}{\infty}=0\)
\(b=\lim n\left(\sqrt{1+\dfrac{2}{n}}-\sqrt{1-\dfrac{2}{n}}-1\right)=+\infty.\left(-1\right)=-\infty\)
\(c=\lim4^n\left(\sqrt{\left(\dfrac{9}{16}\right)^n-\left(\dfrac{3}{16}\right)^n}-1\right)=+\infty.\left(-1\right)=-\infty\)
\(d=\lim n^3\left(3+\dfrac{2}{n}+\dfrac{1}{n^2}\right)=+\infty.3=+\infty\)
20 tháng 11 2023
\(\lim\limits\dfrac{\sqrt{4n^2+1}+2n-1}{\sqrt{n^2+4n+1}+n}\)
\(=\lim\limits\dfrac{\sqrt{4+\dfrac{1}{n^2}}+2-\dfrac{1}{n}}{\sqrt{1+\dfrac{4}{n}+\dfrac{1}{n^2}}+1}=\dfrac{2+2}{1+1}=\dfrac{4}{2}=2\)
\(\lim\limits\left[\sqrt{n}\left(\sqrt{n+1}-n\right)\right]\)
\(=\lim\limits\left[\sqrt{n^2+n}-\sqrt{n^3}\right]\)
\(=\lim\limits\dfrac{n^2+n-n^3}{\sqrt{n^2+n}+\sqrt{n^3}}\)
\(=\lim\limits\dfrac{n^3\left(-1+\dfrac{1}{n}+\dfrac{1}{n^2}\right)}{\sqrt{n^3\left(\dfrac{1}{n}+\dfrac{1}{n^2}\right)}+\sqrt{n^3}}\)
\(=\lim\limits\dfrac{n^3\left(-1+\dfrac{1}{n}+\dfrac{1}{n^2}\right)}{\sqrt{n^3}\left(\sqrt{\dfrac{1}{n}+\dfrac{1}{n^2}}+1\right)}\)
\(=\lim\limits\dfrac{n\sqrt{n}\left(-1+\dfrac{1}{n}+\dfrac{1}{n^2}\right)}{\sqrt{\dfrac{1}{n}+\dfrac{1}{n^2}}+1}\)
\(=-\infty\) vì \(\left\{{}\begin{matrix}lim\left(n\sqrt{n}\right)=+\infty\\lim\left(\dfrac{-1+\dfrac{1}{n}+\dfrac{1}{n^2}}{\sqrt{\dfrac{1}{n}+\dfrac{1}{n^2}}+1}\right)=-\dfrac{1}{1}=-1< 0\end{matrix}\right.\)
