Cho 2017 số nguyên dương \(a_1, a_2, a_3,..., a_{2017}\) thỏa mãn \(\dfrac{1}{a_1}+\dfrac{1}{a_2}+\dfrac{1}{a_3}+...+\dfrac{1}{a_{2017}}=1009\).Chứng minh rằng ít nhất 2 số trong 2017 số nguyên dương đã cho bằng nhau.

Cho \(a, b, c\) là các số hữu tỷ thỏa mãn: \(a+b+c=\dfrac{1}{abc}\). Chứng minh rằng:

\(A=\sqrt{\dfrac{\left(1+b^2c^2\right)\left(1+a^2c^2\right)}{c^2+a^2b^2c^2}}\) là số hữu tỷ

Ta có:

\(\left|x+2\right|>20\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2< -20\\x+2>20\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x< -22\\x>18\end{matrix}\right.\)

\(\left|x-2\right|< 3\)

\(\Leftrightarrow-3< x-2< 3\)

\(\Leftrightarrow-1< x< 5\)

Ta có:

*) \(S=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2015}\)

\(\Rightarrow S=\left(1+\dfrac{1}{3}+...+\dfrac{1}{2015}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2014}\right)\)

\(\Rightarrow S=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2015}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{2014}\right)\)

\(\Rightarrow S=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2015}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{1007}\right)\)

\(\Rightarrow S=\dfrac{1}{1008}+\dfrac{1}{1009}+\dfrac{1}{1010}+...+\dfrac{1}{2015}\)

Vậy \(\left(S-B\right)^{2016}=\left[\left(\dfrac{1}{1008}+\dfrac{1}{1009}+...+\dfrac{1}{2015}\right)-\left(\dfrac{1}{1008}+\dfrac{1}{1009}+...+\dfrac{1}{2015}\right)\right]^{2016}\)

\(\Rightarrow\left(S-B\right)^{2016}=0^{2016}\)

\(\Rightarrow\left(S-B\right)^{2016}=0\)

Cho x; y; z là 3 số thỏa mãn:

\(\dfrac{x}{y}=\dfrac{y}{z}=\dfrac{z}{x}\)

Tìm các số nguyên dương x, y thỏa mãn:

\(\dfrac{3}{x+y}=\dfrac{x}{y+3}=\dfrac{y}{x+3}\)

So sánh: 4³² và 3⁴²

Cho \(\Delta ABC\); hai đường cao BD, CE cắt nhau tại H cho biết AC=BH. Chứng minh rằng: \(\Delta ABC\) có \(\widehat{B}\) \(=45^o\) hoặc \(\widehat{B}=135^o\)

\(\dfrac{\sqrt{\dfrac{9}{4}}-3^{-1}+2018^0}{25\%+1\dfrac{1}{4}-1,3}-\dfrac{\left(\dfrac{-1}{2}\right)^2-\sqrt{\dfrac{4}{9}+0,4}}{0,6-\dfrac{2}{3}\left(\dfrac{-1}{4}-\dfrac{1}{2}\right)}\)

Nhớ giải chi tiết nha

\(\dfrac{\sqrt{\dfrac{9}{4}-3^{-1}+2018^0}}{25\%+1\dfrac{1}{4}-1,3}-\dfrac{\left(\dfrac{-1}{2}\right)^2-\sqrt{\dfrac{4}{9}}+0,4}{0,6-\dfrac{2}{3}.\left(\dfrac{-1}{4}-\dfrac{1}{2}\right)}\)