Chứng minh rằng : \(\sqrt{1+2018^2+\frac{2018^2}{2019^2}}\) +\(\frac{2018}{2019}\)có giá trị là số tự nhiên
chứng minh rằng biểu thức \(B=\sqrt{1+2018^2+\frac{2018^2}{2019^2}}+\frac{2018}{2019}\) có giá trị là 1 số tự nhiên
\(B=\sqrt{\frac{2019^2}{2019^2}+2018^2+\frac{2018^2}{2019^2}}+\frac{2018}{2019}\)
\(B=\sqrt{\frac{\left(2018+1\right)^2}{2019^2}+\frac{2018^2}{2019^2}+2018^2}+\frac{2018}{2019}\)
\(B=\sqrt{\frac{1}{2019^2}+\frac{2018^2+2.2018+2018^2}{2019^2}+2018^2}+\frac{2018}{2019}\)
\(B=\sqrt{\frac{1}{2019^2}+2.2018.\frac{1}{2019}+2018^2}+\frac{2018}{2019}\)
\(B=\sqrt{\left(\frac{1}{2019}+2018\right)^2}+\frac{2018}{2019}\)
\(B=\frac{1}{2019}+2018+\frac{2018}{2019}=2019\) là một số tự nhiên
\(B=\sqrt{1+2018^2+\frac{2018^2}{2019^2}}+\frac{2018}{2019}\)
\(B=\sqrt{1^2+2018^2+\left(-\frac{2018}{2019}\right)^2}+\frac{2018}{2019}\)
\(B=\sqrt{\left(1+2018-\frac{2018}{2019}\right)^2+2.\frac{2018}{2019}+2.\frac{2018^2}{2019}-2.2018}\)\(+\frac{2018}{2019}\)
\(B=\sqrt{\left(1+2018-\frac{2018}{2019}\right)^2+2\left(\frac{2018+2018.2018-2018.2019}{2019}\right)}\)\(+\frac{2018}{2019}\)
\(B=\sqrt{\left(1+2018-\frac{2018}{2019}\right)^2}+\frac{2018}{2019}\)
\(B=1+2018-\frac{2018}{2019}+\frac{2018}{2019}=2019\)
Vậy B có giá trị là 1 số tự nhiên.
S=\(\frac{1}{2018}\left(\frac{2}{1}+\frac{3}{2}+\frac{4}{3}+...+\frac{2019}{2018}\right)\)
Chứng minh S không là số tự nhiên.
1< S< 2
=> S không phải số tự nhiên
1 < S < 2
\(\Rightarrow\) S ko fai là số tự nhiên
Chứng minh rằng biểu thức B = \(\sqrt{1+2017^2+\frac{2017^2}{2018^2}}+\frac{2017}{2018}\) có giá trj là một số tự nhiên
\(B=\sqrt{1+2017^2+\frac{2017^2}{2018^2}}+\frac{2017}{2018}\)
Đặt B = 2017 => B + 1 = 2018
Khi B bằng:
\(B=\sqrt{1+B^2+\frac{B}{\left(B+1\right)^2}}+\frac{B}{B+1}\)
\(B=\sqrt{\frac{\left(B+1\right)^2+B^2\left(B+1\right)^2+B^2}{\left(B+1\right)^2}}+\frac{B}{B+1}\)
\(B=\sqrt{\frac{B^2\left(B+1\right)^2+2B\left(B+1\right)^2+B^2}{\left(B+1\right)^2}}+\frac{B}{B+1}\)
\(B=\sqrt{\frac{\left[B\left(B+1\right)+1\right]^2}{\left(B+1\right)^2}}+\frac{B}{B+1}\)
\(B=\frac{B^2+B+1}{B+1}+\frac{B}{B+1}\left(\text{vi}:a>0\right)\)
\(B=\frac{B^2+2B+1}{B+1}\)
\(B=\frac{\left(B+1\right)^2}{B+1}\)
\(B=B+1\left(\text{vi}:a>0\Rightarrow B+1>0\right)\)
\(B=2017+1\left(\text{vi}:B=2017\right)\)
\(\Rightarrow B=2018\)
Chứng minh rằng : \(2019/ \sqrt[2]{2018} + 2018/\sqrt[2]{2019} > \sqrt[2]{2018} + \sqrt[2]{2019}\)
\(\frac{2019}{\sqrt{2018}}+\frac{2018}{\sqrt{2019}}\ge\frac{\left(\sqrt{2019}+\sqrt{2018}\right)^2}{\sqrt{2018}+\sqrt{2019}}=\sqrt{2018}+\sqrt{2019}\)
Dấu "=" ko xảy ra nên \(\frac{2019}{\sqrt{2018}}+\frac{2018}{\sqrt{2019}}>\sqrt{2018}+\sqrt{2019}\)
Tính giá trị biểu thức \(A=\frac{2^{2018}}{2^{2018}+3^{2019}}+\frac{3^{2019}}{3^{2019}+5^{2000}}+\frac{5^{2000}}{5^{2000}+2^{2018}}\)
Tính \(\sqrt{1+2018^2+\frac{2018^2}{2019^2}}+\frac{2018}{2019}\)
\(1.\)Chứng minh rằng : \(D=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+.....+\frac{2018}{4^{2018}}+\frac{2019}{4^{2019}}< \frac{1}{2}\)
\(D=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+...+\frac{2018}{4^{2018}}+\frac{2019}{4^{2019}}\)
\(\Rightarrow4D=1+\frac{2}{4}+\frac{3}{4^2}+\frac{4}{4^3}+...+\frac{2018}{4^{2017}}+\frac{2019}{4^{2018}}\)
\(\Rightarrow4D-D=1+\frac{2}{4}+\frac{3}{4^2}+\frac{4}{4^3}+...+\frac{2018}{4^{2017}}+\frac{2019}{4^{2018}}\)
\(-\frac{1}{4}-\frac{2}{4^2}-\frac{3}{4^3}-\frac{4}{4^4}-...-\frac{2018}{4^{2018}}-\frac{2019}{4^{2019}}\)
\(\Rightarrow3D=1+\left(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2018}}\right)-\frac{2019}{4^{2019}}\)
Đặt \(M=\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+\frac{1}{4^4}+...+\frac{1}{4^{2018}}\)
\(\Rightarrow4M=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2017}}\)
\(\Rightarrow4M-M=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2017}}\)
\(-\frac{1}{4}-\frac{1}{4^2}-\frac{1}{4^3}-\frac{1}{4^4}-...-\frac{1}{4^{2018}}\)
\(\Rightarrow3M=1-\frac{1}{4^{2018}}\)
\(\Rightarrow M=\frac{1}{3}-\frac{1}{3.4^{2018}}\)
\(\Rightarrow3D=1+\frac{1}{3}-\frac{1}{3.4^{2018}}-\frac{2019}{4^{2019}}\)
\(\Rightarrow3D=\frac{4}{3}-\frac{1}{3.4^{2018}}-\frac{2019}{4^{2019}}< \frac{4}{3}\)
\(\Rightarrow D< \frac{4}{9}=\frac{40}{90}< \frac{45}{90}=\frac{1}{2}\left(đpcm\right)\)
tính \(\sqrt{1+2018^2+\frac{2018}{2019}^2}+\frac{2018}{2019}\)
Đặt \(2018=a\)
\(\Rightarrow\sqrt{1+2018^2+\frac{2018^2}{2019^2}}+\frac{2018}{2019}=\sqrt{1+a^2+\frac{a^2}{\left(a+1\right)^2}}+\frac{a}{a+1}\)
\(=\sqrt{\frac{\left(a^2+a+1\right)^2}{\left(a+1\right)^2}}+\frac{a}{a+1}=\frac{a^2+a+1}{a+1}+\frac{a}{a+1}=\frac{\left(a+1\right)^2}{a+1}=a+1=2019\)
Tính:
\(\sqrt{1+2018^2+\frac{2018^2}{2019^2}}+\frac{2018}{2019}\)