\(\sqrt{1+2017^2+\frac{2017^2}{2018^2}}+\frac{2017}{2018}\)
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DT
2
11 tháng 9 2019
Đặt \(2017=a\)
=>\(2018=a+1\)
Với mọi \(a\in N\) có:\(\sqrt{1+a^2+\frac{a^2}{\left(a+1\right)^2}}=\sqrt{\frac{\left(a+1\right)^2+a^2\left(a+1\right)^2+a^2}{\left(a+1\right)^2}}=\sqrt{\frac{a^2+2a+1+a^2\left(a^2+2a+1\right)+a^2}{\left(a+1\right)^2}}=\sqrt{\frac{2a^2+2a+1+a^4+2a^3+a^2}{\left(a+1\right)^2}}=\sqrt{\frac{\left(a^4+2a^2+1\right)+2a\left(a^2+1\right)+a^2}{\left(a+1\right)^2}}\)
=\(\sqrt{\frac{\left(a^2+1\right)^2+2a\left(a^2+1\right)+a^2}{\left(a+1\right)^2}}=\sqrt{\frac{\left(a^2+a+1\right)}{\left(a+1\right)^2}}=\left|\frac{a^2+a+1}{a+1}\right|\)(do \(a\ge0\))
=\(\frac{a\left(a+1\right)+1}{a+1}=a+\frac{1}{a+1}\)
=> \(\sqrt{1+a^2+\frac{a^2}{\left(a+1\right)^2}}=a+\frac{1}{a+1}\)
Thay a=2017 có:
\(\sqrt{1+2017^2+\left(\frac{2017}{2018}\right)^2}=2017+\frac{1}{2017+1}=2017+\frac{1}{2018}\)
=>\(\sqrt{1+22017^2+\left(\frac{2017}{2018}\right)^2}+\frac{2017}{2018}=2017+\frac{1}{2018}+\frac{2017}{2018}\)
<=> M=2017+1=2018
Vậy M=2018
27 tháng 12 2017
A=\(\frac{2018}{2017^2+1}+\frac{2018}{2017^2+2}+..........+\frac{2018}{2017^2+2017}\)
>\(\frac{2018}{2017^2+2017}+\frac{2018}{2017^2+2017}+........+\frac{2018}{2017^2+2017}\)
\(=\frac{2018}{2017^2+2017}.2017=\frac{2018.2017}{2017\left(2017+1\right)}=1\) (1)
Lại có:A<\(\frac{2018}{2017^2+1}+\frac{2018}{2017^2+1}+.........+\frac{2018}{2017^2+1}\)
\(=\frac{2018}{2017^2+1}.2017=\frac{2018.2017}{2017^2+1}=\frac{2017.\left(2017+1\right)}{2017^2+1}\)
\(=\frac{2017^2+2017}{2017^2+1}=\frac{2017^2+1+2016}{2017^2+1}=1+\frac{2016}{2017^2+1}< 2\) (2)
Từ (1) và (2) suy ra:1 < A < 2
Vậy A không phải là số nguyên
SN
2
2 tháng 9 2017
\(\sqrt{1+a^2+\left(\frac{a}{a+1}\right)^2}\)=\(\sqrt{\frac{\left(a+1\right)^2+a^2.\left(a+1\right)^2+a^2}{\left(a+1\right)^2}}\) =\(\sqrt{\frac{a^2\left(a^2+2a+1+1\right)+\left(a+1\right)^2}{\left(a+1\right)^2}}\)
=\(\sqrt{\frac{a^4+2a^2.\left(a+1\right)+\left(a+1\right)^2}{\left(a+1\right)^2}}\) =\(\sqrt{\frac{\left(a^2+a+1\right)^2}{\left(a+1\right)^2}}=\frac{a^2+a+1}{a+1}=\frac{a\left(a+1\right)+1}{a+1}=a+\frac{1}{a+1}\)
thay vao dau bai ta co
\(2017+\frac{1}{2018}+\frac{2017}{2018}=2017+1=2018\)
LH
22 tháng 5 2020
Câu b đề sai nha, bây giờ đặt \(a=\sqrt{2017},b=\sqrt{2018}\)
Ta có \(\frac{a^2}{b}+\frac{b^2}{a}< a+b\Leftrightarrow ab\left(\frac{a^2}{b}+\frac{b^2}{a}\right)< ab\left(a+b\right)\)
\(\Leftrightarrow a^3+b^3< ab\left(a+b\right)\)(1)
Mà \(ab\left(a+b\right)\le\left(a^2-ab+b^2\right)\left(a+b\right)=a^3+b^3\)(2)
Từ (1), (2) => Sai
22 tháng 5 2020
a) Ta có:
\(\frac{1}{\left(k+1\right)\sqrt{k}}=\frac{k+1-k}{\left(k+1\right)\sqrt{k}}=\frac{\left(\sqrt{k+1}+\sqrt{k}\right)\left(\sqrt{k+1}-\sqrt{k}\right)}{\left(k+1\right)\sqrt{k}}\)\(< \frac{2\sqrt{k+1}\left(\sqrt{k+1}-\sqrt{k}\right)}{\left(k+1\right)\sqrt{k}}=\frac{2\left(\sqrt{k+1}-\sqrt{k}\right)}{\sqrt{k+1}\sqrt{k}}=\frac{2}{\sqrt{k}}-\frac{2}{\sqrt{k+1}}\)
Cho k=1,2,....,n rồi cộng từng vế ta có:
\(\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+....+\frac{1}{\left(n+1\right)\sqrt{n}}< \left(\frac{2}{\sqrt{1}}-\frac{2}{\sqrt{2}}\right)+\left(\frac{2}{\sqrt{2}}-\frac{2}{\sqrt{3}}\right)\)\(+\left(\frac{2}{\sqrt{3}}-\frac{2}{\sqrt{4}}\right)+....+\left(\frac{2}{\sqrt{n}}-\frac{2}{\sqrt{n+1}}\right)=2-\frac{2}{\sqrt{n-1}}< 2\)
16 tháng 10 2018
\(B=\sqrt{1+2017^2+\frac{2017^2}{2018^2}}+\frac{2017}{2018}\)
\(B=\sqrt{\left(1+2.2017+2017^2\right)-2.2017+\frac{2017^2}{2018^2}}+\frac{2017}{2018}\)
\(B=\sqrt{\left(1+2017\right)^2-2.2017+\frac{2017^2}{2018^2}}+\frac{2017}{2018}\)
\(B=\sqrt{2018^2-2.2017+\frac{2017^2}{2018^2}}+\frac{2017}{2018}\)
\(B=\sqrt{\left(2018-\frac{2017}{2018}\right)^2}+\frac{2017}{2018}\)
Mà \(\frac{2017}{2018}< 1\Rightarrow2018-\frac{2017}{2018}>0\)
\(\Rightarrow B=2018-\frac{2017}{2018}+\frac{2017}{2018}\)
\(B=2018\)
Vậy bt B có giá trị nguyên
31 tháng 3 2018
\(B=\sqrt{1+2017^2+\frac{2017^2}{2018^2}}+\frac{2017}{2018}\)
Đặt B = 2017 => B + 1 = 2018
Khi B bằng:
\(B=\sqrt{1+B^2+\frac{B}{\left(B+1\right)^2}}+\frac{B}{B+1}\)
\(B=\sqrt{\frac{\left(B+1\right)^2+B^2\left(B+1\right)^2+B^2}{\left(B+1\right)^2}}+\frac{B}{B+1}\)
\(B=\sqrt{\frac{B^2\left(B+1\right)^2+2B\left(B+1\right)^2+B^2}{\left(B+1\right)^2}}+\frac{B}{B+1}\)
\(B=\sqrt{\frac{\left[B\left(B+1\right)+1\right]^2}{\left(B+1\right)^2}}+\frac{B}{B+1}\)
\(B=\frac{B^2+B+1}{B+1}+\frac{B}{B+1}\left(\text{vi}:a>0\right)\)
\(B=\frac{B^2+2B+1}{B+1}\)
\(B=\frac{\left(B+1\right)^2}{B+1}\)
\(B=B+1\left(\text{vi}:a>0\Rightarrow B+1>0\right)\)
\(B=2017+1\left(\text{vi}:B=2017\right)\)
\(\Rightarrow B=2018\)
Lời giải:
\(\sqrt{1+2017^2+\frac{2017^2}{2018^2}}+\frac{2017}{2018}\)
\(=\sqrt{(1+2017)^2-2.2017+\frac{2017^2}{2018^2}}+\frac{2017}{2018}\)
\(\sqrt{2018^2-2.2018.\frac{2017}{2018}+\frac{2017^2}{2018^2}}+\frac{2017}{2018}\)
\(=\sqrt{(2018-\frac{2017}{2018})^2}+\frac{2017}{2018}=2018-\frac{2017}{2018}+\frac{2017}{2018}=2018\)