Chứng minh rằng : \(\sqrt{1+2018^2+\frac{2018^2}{2019^2}}\) +\(\frac{2018}{2019}\)có giá trị là số tự nhiên
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Những câu hỏi liên quan
4 tháng 6 2019
\(B=\sqrt{\frac{2019^2}{2019^2}+2018^2+\frac{2018^2}{2019^2}}+\frac{2018}{2019}\)
\(B=\sqrt{\frac{\left(2018+1\right)^2}{2019^2}+\frac{2018^2}{2019^2}+2018^2}+\frac{2018}{2019}\)
\(B=\sqrt{\frac{1}{2019^2}+\frac{2018^2+2.2018+2018^2}{2019^2}+2018^2}+\frac{2018}{2019}\)
\(B=\sqrt{\frac{1}{2019^2}+2.2018.\frac{1}{2019}+2018^2}+\frac{2018}{2019}\)
\(B=\sqrt{\left(\frac{1}{2019}+2018\right)^2}+\frac{2018}{2019}\)
\(B=\frac{1}{2019}+2018+\frac{2018}{2019}=2019\) là một số tự nhiên
4 tháng 6 2019
\(B=\sqrt{1+2018^2+\frac{2018^2}{2019^2}}+\frac{2018}{2019}\)
\(B=\sqrt{1^2+2018^2+\left(-\frac{2018}{2019}\right)^2}+\frac{2018}{2019}\)
\(B=\sqrt{\left(1+2018-\frac{2018}{2019}\right)^2+2.\frac{2018}{2019}+2.\frac{2018^2}{2019}-2.2018}\)\(+\frac{2018}{2019}\)
\(B=\sqrt{\left(1+2018-\frac{2018}{2019}\right)^2+2\left(\frac{2018+2018.2018-2018.2019}{2019}\right)}\)\(+\frac{2018}{2019}\)
\(B=\sqrt{\left(1+2018-\frac{2018}{2019}\right)^2}+\frac{2018}{2019}\)
\(B=1+2018-\frac{2018}{2019}+\frac{2018}{2019}=2019\)
Vậy B có giá trị là 1 số tự nhiên.
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:
\(B=\sqrt{1+B^2+\frac{B}{\left(B+1\right)^2}}+\frac{B}{B+1}\)
\(B=\sqrt{\frac{\left(B+1\right)^2+B^2\left(B+1\right)^2+B^2}{\left(B+1\right)^2}}+\frac{B}{B+1}\)
\(B=\sqrt{\frac{B^2\left(B+1\right)^2+2B\left(B+1\right)^2+B^2}{\left(B+1\right)^2}}+\frac{B}{B+1}\)
\(B=\sqrt{\frac{\left[B\left(B+1\right)+1\right]^2}{\left(B+1\right)^2}}+\frac{B}{B+1}\)
\(B=\frac{B^2+B+1}{B+1}+\frac{B}{B+1}\left(\text{vi}:a>0\right)\)
\(B=\frac{B^2+2B+1}{B+1}\)
\(B=\frac{\left(B+1\right)^2}{B+1}\)
\(B=B+1\left(\text{vi}:a>0\Rightarrow B+1>0\right)\)
\(B=2017+1\left(\text{vi}:B=2017\right)\)
\(\Rightarrow B=2018\)
DG
1
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 9 2020
\(\frac{2019}{\sqrt{2018}}+\frac{2018}{\sqrt{2019}}\ge\frac{\left(\sqrt{2019}+\sqrt{2018}\right)^2}{\sqrt{2018}+\sqrt{2019}}=\sqrt{2018}+\sqrt{2019}\)
Dấu "=" ko xảy ra nên \(\frac{2019}{\sqrt{2018}}+\frac{2018}{\sqrt{2019}}>\sqrt{2018}+\sqrt{2019}\)
GN
0
2T
14 tháng 8 2019
\(D=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+...+\frac{2018}{4^{2018}}+\frac{2019}{4^{2019}}\)
\(\Rightarrow4D=1+\frac{2}{4}+\frac{3}{4^2}+\frac{4}{4^3}+...+\frac{2018}{4^{2017}}+\frac{2019}{4^{2018}}\)
\(\Rightarrow4D-D=1+\frac{2}{4}+\frac{3}{4^2}+\frac{4}{4^3}+...+\frac{2018}{4^{2017}}+\frac{2019}{4^{2018}}\)
\(-\frac{1}{4}-\frac{2}{4^2}-\frac{3}{4^3}-\frac{4}{4^4}-...-\frac{2018}{4^{2018}}-\frac{2019}{4^{2019}}\)
\(\Rightarrow3D=1+\left(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2018}}\right)-\frac{2019}{4^{2019}}\)
Đặt \(M=\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+\frac{1}{4^4}+...+\frac{1}{4^{2018}}\)
\(\Rightarrow4M=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2017}}\)
\(\Rightarrow4M-M=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2017}}\)
\(-\frac{1}{4}-\frac{1}{4^2}-\frac{1}{4^3}-\frac{1}{4^4}-...-\frac{1}{4^{2018}}\)
\(\Rightarrow3M=1-\frac{1}{4^{2018}}\)
\(\Rightarrow M=\frac{1}{3}-\frac{1}{3.4^{2018}}\)
\(\Rightarrow3D=1+\frac{1}{3}-\frac{1}{3.4^{2018}}-\frac{2019}{4^{2019}}\)
\(\Rightarrow3D=\frac{4}{3}-\frac{1}{3.4^{2018}}-\frac{2019}{4^{2019}}< \frac{4}{3}\)
\(\Rightarrow D< \frac{4}{9}=\frac{40}{90}< \frac{45}{90}=\frac{1}{2}\left(đpcm\right)\)
N
1
24 tháng 9 2018
Đặt \(2018=a\)
\(\Rightarrow\sqrt{1+2018^2+\frac{2018^2}{2019^2}}+\frac{2018}{2019}=\sqrt{1+a^2+\frac{a^2}{\left(a+1\right)^2}}+\frac{a}{a+1}\)
\(=\sqrt{\frac{\left(a^2+a+1\right)^2}{\left(a+1\right)^2}}+\frac{a}{a+1}=\frac{a^2+a+1}{a+1}+\frac{a}{a+1}=\frac{\left(a+1\right)^2}{a+1}=a+1=2019\)
NT
0
Căn bậc 2 của 1 là 1,của 2018 bình phương là 2018,2018 bình phương/2019 bình phương là 2018/2019 nên cái căn đó có giá trị là 1+2018+2018/2019 nha.bn lấy 2018/2019+2018/2019 nếu là số tự nhiên thì biểu thức này là STN
\(\sqrt{1+2018^2+\frac{2018^2}{2019^2}}+\frac{2018}{2019}\)
\(=\)\(\sqrt{\left(1+2.2018+2018^2\right)-2.2018+\frac{2018^2}{2019^2}}+\frac{2018}{2019}\)
\(=\)\(\sqrt{2019^2-2.2018+\frac{2018^2}{2019^2}}+\frac{2018}{2019}\)
\(=\)\(\sqrt{\left(2019-\frac{2018}{2019}\right)^2}+\frac{2018}{2019}\)
\(=\)\(\left|2019-\frac{2018}{2019}\right|+\frac{2018}{2019}=2019-\frac{2018}{2019}+\frac{2018}{2019}=2019\)
\(\Rightarrow\)\(\sqrt{1+2018^2+\frac{2018^2}{2019^2}}+\frac{2018}{2019}\) là số tự nhiên ( đpcm )
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