trong bai : cho a= \(\dfrac{1}{2\sqrt{1} 1\sqrt{2}} \dfrac{1}{3\sqrt{2} 2\sqrt{3}} \dfrac{1}{4\sqrt{3} 3\sqrt{4}} .... \dfrac{1}{\left(n 1\right)\sqrt{n} n\sqrt{n 1}}< 1\) co phan huong dan : \(\dfr...

NV

Nguyễn Viết Thắng

5 tháng 7 2017

trong bai : cho a= \(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+....+\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}} 1\) co phan huong dan : \(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{1}{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n+1}+\sqrt{n}\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\) cho minh hoi buoc : \(\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}}\) tu dau ra .( giai thich...

Đọc tiếp

#Toán lớp 9

1

TT

Trần Thiên Kim

10 tháng 9 2017

\(\dfrac{1}{\sqrt{n}.\sqrt{n+1}.\left(\sqrt{n+1}+\sqrt{n}\right)}=\dfrac{1}{\sqrt{n}.\sqrt{n+1}}.\dfrac{1}{\sqrt{n+1}+\sqrt{n}}=\dfrac{1}{\sqrt{n}.\sqrt{n+1}}.\dfrac{\sqrt{n+1}-\sqrt{n}}{n+1-n}=\dfrac{1}{\sqrt{n}.\sqrt{n+1}}.\left(\sqrt{n+1}-\sqrt{n}\right)=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}}\)

Đúng(0)

DD

Đặng Dung

8 tháng 10 2017

1) Chứng minh rằng: \(1+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{3\sqrt{3}}+...+\dfrac{1}{n\sqrt{n}} 2\sqrt{2}\left(n\in N\right)\) 2) Chứng minh rằng: \(\dfrac{2}{3}+\sqrt{n+1} 1+\sqrt{2}+\sqrt{3}+...+\sqrt{n} \dfrac{2}{3}\left(n+1\right)\sqrt{n}\) 3) \(2\sqrt{n}-3 \dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{n}} 2\sqrt{n}-2\) 4) \(\dfrac{\sqrt{2}-\sqrt{1}}{2+1}+\dfrac{\sqrt{3}-\sqrt{2}}{3+2}+...+\dfrac{\sqrt{n+1}-\sqrt{n}}{n+1+n}...

Đọc tiếp

#Toán lớp 9

0

NP

Nguyễn Phạm Hoàng trang

25 tháng 9 2017

CMR:

\(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{3}}+....+\dfrac{1}{\left(n+1\right)\left(\sqrt{n}+n\sqrt{n+1}\right)}< 1\)

#Toán lớp 9

0

....

30 tháng 6 2021

chứng minh rằng với số tự nhiên n,n lớn hơn 4 ta có:

\(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+...+\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}< 1\)

#Toán lớp 9

1

NV

Nguyễn Việt Lâm

Giáo viên

30 tháng 6 2021

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

\(=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)

Do đó:

\(VT=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)

\(VT=1-\dfrac{1}{\sqrt{n+1}}< 1\) (đpcm)

Đúng(2)

PT

Phương Thảo

7 tháng 7 2017

Rút gọn: A = \(\dfrac{4+\sqrt{7}}{3\sqrt{2}+\sqrt{4+\sqrt{7}}}+\dfrac{4-\sqrt{7}}{3\sqrt{2}-\sqrt{4-\sqrt{7}}}\) B = \(\dfrac{3\sqrt{2}+\sqrt{11}}{\sqrt{2}+\sqrt{6+\sqrt{11}}}+\dfrac{3\sqrt{2}-\sqrt{11}}{\sqrt{2}-\sqrt{6-\sqrt{11}}}+18\) C = \(\dfrac{1}{\sqrt{3}+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{7}}+...+\dfrac{1}{\sqrt{2n+1}+\sqrt{2n+3}}\)với n thuộc N* D =...

Đọc tiếp

#Toán lớp 9

0

TN

Thư Nguyễn Huỳnh Anh

20 tháng 1 2021

...

Đọc tiếp

#Toán lớp 11

2

NV

Nguyễn Việt Lâm

Giáo viên

20 tháng 1 2021

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

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

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

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

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

Đúng(1)

NV

Nguyễn Việt Lâm

Giáo viên

20 tháng 1 2021

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

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

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

Đúng(0)

HG

Hoàng Giang

25 tháng 12 2023

a, \(\left(18\dfrac{1}{3}:\sqrt{225}+8\dfrac{2}{3}.\sqrt{\dfrac{49}{4}}\right)\): \(\left[\left(12\dfrac{1}{3}+8\dfrac{6}{7}\right)-\dfrac{\left(\sqrt{7}\right)^2}{\left(3\sqrt{2}\right)^2}\right]\): \(\dfrac{1704}{445}\)b, \(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+...+\(\dfrac{1}{99.100}\)c, \(\left(1-\dfrac{1}{2}\right)\)x\(\left(1-\dfrac{1}{3}\right)\)x.....x\(\left(1-\dfrac{1}{n+1}\right)\) (n ϵ N)d, -66 x \(\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{11}\right)\) + 124 x -37 + 63 x...

Đọc tiếp

#Toán lớp 7

1

NL

Nguyễn Lê Phước Thịnh

31 tháng 12 2023

a: \(\left(18\dfrac{1}{3}:\sqrt{225}+8\dfrac{2}{3}\cdot\sqrt{\dfrac{49}{4}}\right):\left[\left(12\dfrac{1}{3}+8\dfrac{6}{7}\right)-\dfrac{\left(\sqrt{7}\right)^2}{\left(3\sqrt{2}\right)^2}\right]:\dfrac{1704}{445}\)

\(=\left(\dfrac{55}{3}:15+\dfrac{26}{3}\cdot\dfrac{7}{4}\right):\left[\left(12+\dfrac{1}{3}+8+\dfrac{6}{7}\right)-\dfrac{7}{18}\right]\cdot\dfrac{445}{1704}\)

\(=\left(\dfrac{55}{45}+\dfrac{91}{6}\right):\left[20+\dfrac{101}{126}\right]\cdot\dfrac{445}{1704}\)

\(=\dfrac{295}{18}:\dfrac{2621}{126}\cdot\dfrac{445}{1704}\)

\(=\dfrac{295}{18}\cdot\dfrac{126}{2621}\cdot\dfrac{445}{1704}\simeq0,21\)

b: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)

c: \(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{n+1}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{n}{n+1}\)

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

d: \(-66\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{11}\right)+124\cdot\left(-37\right)+63\cdot\left(-124\right)\)

\(=-66\cdot\dfrac{33-22+6}{66}+124\left(-37-63\right)\)

\(=-17-12400=-12417\)

e: \(\dfrac{7}{4}\left(\dfrac{33}{12}+\dfrac{3333}{2020}+\dfrac{333333}{303030}+\dfrac{33333333}{42424242}\right)\)

\(=\dfrac{7}{4}\left(\dfrac{33}{12}+\dfrac{33}{20}+\dfrac{33}{30}+\dfrac{33}{42}\right)\)

\(=\dfrac{7}{4}\cdot33\cdot\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\right)\)

\(=33\cdot\dfrac{7}{4}\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\right)\)

\(=33\cdot\dfrac{7}{4}\cdot\left(\dfrac{1}{3}-\dfrac{1}{7}\right)\)

\(=33\cdot\dfrac{7}{4}\cdot\dfrac{4}{21}=\dfrac{33\cdot1}{3}=11\)

Đúng(0)

NM

Nguyễn Mary

23 tháng 5 2018

Bài 1: CM với mọi số nguyên dương n thì \(\dfrac{1}{2\sqrt{1}}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{4\sqrt{3}}+.....+\dfrac{1}{\left(n+1\right)\sqrt{n}}< 2\)

Bài 2: CM với mọi số tự nhiên n>=2 đều có \(\sqrt{n}< \dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{4}}+.....+\dfrac{1}{\sqrt{n}}< 2\sqrt{n}\)

#Toán lớp 9

1

N

ngonhuminh

24 tháng 5 2018

bai 1

(n+1)√n=√n^3+√n>2√(n^3.n)=2n^2>2(n^2-1)=2(n-1)(n+1)

1/[(n+1)√n]<1/[2(n-1)(n+1)]=1/4.[2/(n-1)(n+1)]

A=..

n =1 yes

n>1

A<1+1/4[2/1.3+2/3.5+..+2/(n-1)(n+1)

A<1+1/4[ 2-1/(n+1)]<1+1/2<2=>dpcm

Đúng(0)

DN

đoàn ngọc hân

17 tháng 1 2021

Tính các giới hạn sau

1,Lim\(\left(\dfrac{2n^3}{2n^2+3}+\dfrac{1-5n^2}{5n+1}\right)\)

2,a,Lim\(\left(\sqrt{n^2+n}-\sqrt{n^2+2}\right)\)

b,Lim\(\dfrac{\sqrt{n^4+3n-2}}{2n^2-n+3}\)

c,Lim\(\dfrac{\sqrt{n^2-4n}-\sqrt{4n^2+1}}{\sqrt{3n^2+1}-n}\)

#Toán lớp 11

2

NV

Nguyễn Việt Lâm

Giáo viên

17 tháng 1 2021

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

\(=\lim\left(\dfrac{2n^3-13n^2+3}{\left(2n^2+3\right)\left(5n+1\right)}\right)=\lim\dfrac{2-\dfrac{13}{n}+\dfrac{3}{n^3}}{\left(2+\dfrac{3}{n^2}\right)\left(5+\dfrac{1}{n}\right)}=\dfrac{2}{2.5}=\dfrac{1}{5}\)

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

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

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

Đúng(1)

DT

Dương thị bầu

15 tháng 3 2022

Lim 3.4n-2.13n/5n+6.13n

Đúng(0)

TK

Trần Kiều Thi

17 tháng 6 2018