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\(B=\sqrt{\frac{2019^2}{2019^2}+2018^2+\frac{2018^2}{2019^2}}+\frac{2018}{2019}\)
\(B=\sqrt{\frac{\left(2018+1\right)^2}{2019^2}+\frac{2018^2}{2019^2}+2018^2}+\frac{2018}{2019}\)
\(B=\sqrt{\frac{1}{2019^2}+\frac{2018^2+2.2018+2018^2}{2019^2}+2018^2}+\frac{2018}{2019}\)
\(B=\sqrt{\frac{1}{2019^2}+2.2018.\frac{1}{2019}+2018^2}+\frac{2018}{2019}\)
\(B=\sqrt{\left(\frac{1}{2019}+2018\right)^2}+\frac{2018}{2019}\)
\(B=\frac{1}{2019}+2018+\frac{2018}{2019}=2019\) 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.
\(B=\sqrt{1+2017^2+\frac{2017^2}{2018^2}}+\frac{2017}{2018}\)
Đặt B = 2017 => B + 1 = 2018
Khi B bằng:
\(B=\sqrt{1+B^2+\frac{B}{\left(B+1\right)^2}}+\frac{B}{B+1}\)
\(B=\sqrt{\frac{\left(B+1\right)^2+B^2\left(B+1\right)^2+B^2}{\left(B+1\right)^2}}+\frac{B}{B+1}\)
\(B=\sqrt{\frac{B^2\left(B+1\right)^2+2B\left(B+1\right)^2+B^2}{\left(B+1\right)^2}}+\frac{B}{B+1}\)
\(B=\sqrt{\frac{\left[B\left(B+1\right)+1\right]^2}{\left(B+1\right)^2}}+\frac{B}{B+1}\)
\(B=\frac{B^2+B+1}{B+1}+\frac{B}{B+1}\left(\text{vi}:a>0\right)\)
\(B=\frac{B^2+2B+1}{B+1}\)
\(B=\frac{\left(B+1\right)^2}{B+1}\)
\(B=B+1\left(\text{vi}:a>0\Rightarrow B+1>0\right)\)
\(B=2017+1\left(\text{vi}:B=2017\right)\)
\(\Rightarrow B=2018\)
Vì a ; b ; c dương \(\Rightarrow a+b+c\ne0\)
Ta có : \(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ac=0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow a-b=0;b-c=0;c-a=0\Leftrightarrow a=b=c\)
Vậy \(A=\left(1-\frac{a}{b}\right)\left(2018-\frac{b}{c}\right)\left(2019-\frac{c}{a}\right)=\left(1-1\right).\left(...\right)=0\)
Ta có: \(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}=\left(1+\frac{1}{2018}\right)+\left(\frac{1}{2}+\frac{1}{2017}\right)+...+\left(\frac{1}{1009}+\frac{1}{1010}\right)\)
\(=\frac{2019}{1.2018}+\frac{2019}{2.2017}+...+\frac{2019}{1009.1010}\)
\(=2019\left(\frac{1}{1.2018}+\frac{1}{2.2017}+...+\frac{1}{1009.1010}\right)\)
Do đó \(A=1.2.3....2018.2019\left(\frac{1}{1.2018}+\frac{1}{2.2017}+...+\frac{1}{1009.1010}\right)⋮2019\) (đpcm)