Tính :\(P=\sqrt{1+1999^2+\dfrac{1999^2}{2000^2}}+\dfrac{1999}{2000}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
NT
1
29 tháng 10 2018
\(P=\sqrt{1+1999^2+\dfrac{1999^2}{2000^2}}+\dfrac{1999}{2000}=\sqrt{\dfrac{2000^2+1999^2.2000^2+1999^2}{2000^2}}+\dfrac{1999}{2000}=\dfrac{\sqrt{2000^2+\left(2000-1\right)^2.2000^2+1999^2}}{2000}+\dfrac{1999}{2000}=\dfrac{\sqrt{2000^2+\left(2000^2-2.2000+1\right).2000^2+1999^2}+1999}{2000}=\dfrac{\sqrt{2000^2+2000^4-2.2000.2000^2+2000^2+1999^2}+1999}{2000}=\dfrac{\sqrt{2000^4+2.2000^2-2.\left(1999+1\right).2000^2+1999^2}+1999}{2000}=\dfrac{\sqrt{2000^4+2.2000^2-2.1999.2000^2-2.2000^2+1999^2}+1999}{2000}=\dfrac{\sqrt{2000^4-2.1999.2000^2+1999^2}+1999}{2000}=\dfrac{\sqrt{\left(2000^2-1999\right)^2}+1999}{2000}=\dfrac{2000^2-1999+1999}{2000}=\dfrac{2000^2}{2000}=2000\)
7 tháng 8 2017
Ta chứng minh công thức:
\(1+\dfrac{1}{n^2}+\dfrac{1}{\left(n+1\right)^2}=\left(1+\dfrac{1}{n}+\dfrac{1}{n+1}\right)^2\) bằng cách quy đồng biểu thức ở vế phải rồi áp dụng vào bài tập
VD
3
3 tháng 11 2016
Đặt 2000 = a thì ta có
A = \(\sqrt{1+\left(a-1\right)^2+\frac{\left(a-1\right)^2}{a^2}}+\frac{a-1}{a}\)
\(=\sqrt{\frac{a^4-2a^3+3a^2-2a+1}{a^2}}+\frac{a-1}{a}\)
\(=\frac{a^2-a+1}{a}+\frac{a-1}{a}=a=2000\)
2 tháng 9 2018
Chứng minh công thức:
\(\sqrt{1+\dfrac{1}{a^2}+\dfrac{1}{\left(a+1\right)^2}}\)
\(=\sqrt{\dfrac{a^2\left(a+1\right)^2+\left(a+1\right)^2+a^2}{a^2\left(a+1\right)^2}}\)
\(=\sqrt{\dfrac{a^2\left(a^2+2a+1\right)+a^2+2a+1+a^2}{a^2\left(a+1\right)^2}}\)
\(=\sqrt{\dfrac{a^4+2a^3+a^2+a^2+2a+1+a^2}{a^2\left(a+1\right)^2}}\)
\(=\sqrt{\dfrac{a^4+2a^3+3a^2+2a+1}{a^2\left(a+1\right)^2}}\)
=\(\sqrt{\dfrac{\left(a^2\right)^2+2a^2a+2a^2+2a+a^2+1}{a^2\left(a+1\right)^2}}\)
\(=\sqrt{\dfrac{\left(a^2+a+1\right)^2}{a^2\left(a+1\right)^2}}\)
\(=\dfrac{a^2+a+1}{a\left(a+1\right)}\)
\(=\dfrac{a\left(a+1\right)+1}{a\left(a+1\right)}\)
\(=1+\dfrac{1}{a\left(a+1\right)}\)
\(=1+\dfrac{1}{a}-\dfrac{1}{a+1}\)
Áp dụng công thức ta có:
\(C=\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+...+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{1999^2}+\dfrac{1}{2000^2}}\)
\(=1+\dfrac{1}{2}-\dfrac{1}{3}+...+1+\dfrac{1}{1999}-\dfrac{1}{2000}\)
\(=2000-\dfrac{1}{2000}=\dfrac{1999}{2000}\)
VN
Võ Ngọc Phương
VIP
3 tháng 8 2023
So sánh
\(A=\dfrac{1999^{1999}+1}{1999^{1998}+1}\) ; \(B=\dfrac{1999^{2000}+1}{1999^{1999}+1}\)
Ta có: \(B=\dfrac{1999^{2000}+1}{1999^{1999}+1}>1\) ( vì tử > mẫu )
Do đó: \(B=\dfrac{1999^{2000}+1}{1999^{1999}+1}>\dfrac{1999^{2000}+1+1998}{1999^{1999}+1+1998}=\dfrac{1999^{2000}+1999}{1999^{1999}+1999}=\dfrac{1999.\left(1999^{1999}+1\right)}{1999.\left(1999^{1998}+1\right)}=\dfrac{1999^{1999}+1}{1999^{1998}+1}=A\)
Vậy B > A
Chúc bạn học tốt
BB
2
ML
14 tháng 7 2015
Với số nguyên dương n, ta có:
\(1+n^2+\left(\frac{n}{n+1}\right)^2=\frac{\left(n+1\right)^2+n^2\left(n+1\right)^2+n^2}{\left(n+1\right)^2}=\frac{n^2+2n+1+n^2+n^2\left(n+1\right)^2}{\left(n+1\right)^2}\)
\(=\frac{n^2\left(n+1\right)^2+2n\left(n+1\right)+1}{\left(n+1\right)^2}=\frac{\left[n\left(n+1\right)+1\right]^2}{\left(n+1\right)^2}=\left(\frac{n^2+n+1}{n+1}\right)^2\)
\(\Rightarrow\sqrt{1+n^2+\left(\frac{n}{n+1}\right)^2}=\frac{n^2+n+1}{n+1}=n+\frac{1}{n+1}\)
\(\Rightarrow P=\left(1999+\frac{1}{2000}\right)+\frac{1999}{2000}=1999+1=2000\)
5 tháng 12 2018
Cách ez hđt lp 8 nhé
\(P=\sqrt{\left(1+2.1999+1999^2\right)-2.1999+\frac{1999^2}{2000^2}}+\frac{1999}{2000}\)
\(P=\sqrt{\left(1+1999\right)^2-2.1999+\frac{1999^2}{2000^2}}+\frac{1999}{2000}\)
\(P=\sqrt{2000^2-2.1999+\frac{1999^2}{2000^2}}+\frac{1999}{2000}\)
\(P=\sqrt{\left(2000-\frac{1999}{2000}\right)^2}+\frac{1999}{2000}\)
\(P=\left|2000-\frac{1999}{2000}\right|+\frac{1999}{2000}=2000-\frac{1999}{2000}+\frac{1999}{2000}=2000\)
...
LD
18 tháng 1 2019
Ta có:
\(A-B=\dfrac{1999^{1999}+1}{1999^{1998}+1}-\dfrac{1999^{2000}+1}{1999^{1999}+1}\)
\(=\dfrac{\left(1999^{1999}+1\right)^2-\left(1999^{1998}+1\right)\left(1999^{2000}+1\right)}{\left(1999^{1998}+1\right)\left(1999^{1999}+1\right)}\)
\(=\dfrac{1999^{3998}+2\cdot1999^{1999}+1-\left(1999^{3998}+1999^{1998}+1999^{2000}+1\right)}{1999^{3997}+1999^{1998}+1999^{1999}+1}\)
\(=\dfrac{2\cdot1999^{1999}-1999^{1998}-1999^{2000}}{1999^{3997}+1999^{1998}+1999^{1999}+1}\)
Mà \(2\cdot1999^{1999}-1999^{1998}-1999^{2000}=-\left[\left(1999^{999}\right)^2-2\cdot1999^{999}\cdot1999^{1000}+\left(1999^{1000}\right)^2\right]\)
\(=-\left(1999^{999}-1999^{1000}\right)^2< 0\)
Mà mẫu số > 0
\(\Rightarrow A-B< 0\Leftrightarrow A< B\)
28 tháng 1 2019
A=\(\dfrac{1999^{1999}+1999-1998}{1999^{1998}+1}\) B=\(\dfrac{1999^{2000}+1999-1998}{1999^{1999}+1}\)
A=1999-\(\dfrac{1998}{1999^{1998}+1}\) B=1999-\(\dfrac{1998}{1999^{1999}+1}\)
Vì 19991998+1<19991999+1 nên
\(\dfrac{1}{1999^{1998}+1}\)>\(\dfrac{1}{1999^{1999}+1}\) nên \(\dfrac{-1}{1999^{1998}+1}< \dfrac{-1}{1999^{1999}+1}\)
A=1999+\(\dfrac{-1}{1999^{1998}+1}< 1999+\dfrac{-1}{1999^{1999}+1}\)=B
A<B
DP
0
\(\sqrt{1+a^2+\dfrac{a^2}{\left(a+1\right)^2}}\)
\(=\sqrt{1^2+a^2+\left(\dfrac{a}{a+1}\right)^2+2a-\dfrac{2a}{a+1}-\dfrac{2a^2}{a+1}}\)
(vì \(2a-\dfrac{2a}{a+1}-\dfrac{2a^2}{a+1}=\dfrac{2a^2+2a-2a-2a^2}{a+1}=0\))
\(=\sqrt{\left(1+a-\dfrac{a}{a+1}\right)^2}\)
\(=\left|1+a-\dfrac{a}{a+1}\right|\)
Áp dụng vào P, ta có:
\(P=\sqrt{1+1999^2+\dfrac{1999^2}{2000^2}}+\dfrac{1999}{2000}\)
\(=\left|1+1999-\dfrac{1999}{2000}\right|+\dfrac{1999}{2000}\)
\(=2000\)