ko biết

ai bt giúp mik cái

\(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}...\frac{2018}{2019}\)

\(=\frac{1\cdot2\cdot3\cdot4\cdot\cdot\cdot2018}{2\cdot3\cdot4\cdot5\cdot\cdot\cdot2019}\)

\(=\frac{1\cdot\left(2\cdot3\cdot4\cdot\cdot\cdot2018\right)}{\left(2\cdot4\cdot5\cdot\cdot\cdot2018\right)\cdot2019}\)

\(=\frac{1}{2019}\)

Vậy .......................................

1/2.2/3.3/4.....2018/2019

=1.2.3......2018/2.3.4......2019

=1/2019

\(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2018}{2019}\)

\(=\frac{1.2.3...2018}{2.3.4...2019}\)

\(=\frac{1}{2019}\)

\(A=\dfrac{2020}{2019}-\dfrac{2019}{2018}+\dfrac{1}{2018\times2019}\)

\(A=\dfrac{2020}{2019}-\dfrac{2019}{2018}+\dfrac{1}{2018}-\dfrac{1}{2019}\)

\(A=\left(\dfrac{2020}{2019}-\dfrac{1}{2019}\right)-\left(\dfrac{2019}{2018}-\dfrac{1}{2018}\right)\)

\(A=\left(\dfrac{2020-1}{2019}\right)-\left(\dfrac{2019-1}{2018}\right)\)

\(A=1-1\)

\(A=0.\)

\(A=\dfrac{2020}{2019}-\dfrac{2019}{2018}+\dfrac{1}{2018\times2019}\)

\(A=\dfrac{2020}{2019}-\dfrac{2019}{2018}+\dfrac{1}{2018}-\dfrac{1}{2019}\)

\(A=\left(\dfrac{2020}{2019}-\dfrac{1}{2019}\right)-\left(\dfrac{2019}{2018}-\dfrac{1}{2018}\right)\)

\(A=\dfrac{2019}{2019}-\dfrac{2018}{2018}\)

\(A=1-1\)

\(A=0\)

M= ( 1/1+1/2+1/3+...+1/2018).(673.3).2.4.5....2018

M= (1/1+1/2+1/3+...+1/2018).2019.2.4.5...2018

vi bieu thuc tren co so 2019

=> M chia het cho 2019

nhầm đề r NGUYỄN BÙI KHÁNH NGỌC ạ

Mk giúp bn bài 1 thui đc ko 

Bài dưới mk ko bít ~~~~

Sao các bn cứ tk sai mk vô cớ thế nhỉ , mk đã lm j sai , mk chỉ nói là mk ko bít bài 2 thui mak tự nhiên tk ngta sai , bn nào tk mk sai rồi các bn sẽ biết hậu quả thôi :PPP

YuKo à , trước khi gửi câu hỏi phải gửi đoàng hoàng , nếu muốn nói là ko bt làm bài 2 thì gửi tin nhắn nha (rút kinh nghiệm lần sau)

\(P=\dfrac{x}{\sqrt{y}}+\dfrac{y}{\sqrt{x}}\Rightarrow P^2=\dfrac{x^2}{y}+\dfrac{y^2}{x}+2\sqrt{xy}\)

\(P^2=\left(\dfrac{x^2}{y}+\sqrt{xy}+\sqrt{xy}\right)+\left(\dfrac{y^2}{x}+\sqrt{xy}+\sqrt{xy}\right)-2\sqrt{xy}\)

\(P^2\ge3x+3y-2\sqrt{xy}\ge3\left(x+y\right)-\left(x+y\right)=2\left(x+y\right)=4038\)

\(\Rightarrow P\ge\sqrt{4038}\)

Dấu "=" xảy ra khi \(x=y=\dfrac{2019}{2}\)

Ta có:

\(P=\dfrac{x}{\sqrt{2019-x}}+\dfrac{y}{\sqrt{y-2019}}=\dfrac{x}{\sqrt{y}}+\dfrac{y}{\sqrt{x}}\ge\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^2}{\sqrt{x}+\sqrt{y}}=\sqrt{x}+\sqrt{y}\)

Lại có:

\(P=\dfrac{x}{\sqrt{2019-x}}+\dfrac{y}{\sqrt{2019-y}}=\dfrac{2019-y}{\sqrt{y}}+\dfrac{2019-x}{\sqrt{x}}\\ =\dfrac{2019}{\sqrt{x}}+\dfrac{2019}{\sqrt{y}}-\sqrt{x}-\sqrt{y}\)

\(\Rightarrow2P=\dfrac{2019}{\sqrt{x}}+\dfrac{2019}{\sqrt{y}}=2019\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\right)\ge2019\cdot\dfrac{2}{\sqrt[4]{xy}}\\ \ge2019\dfrac{2}{\sqrt[2]{\dfrac{x+y}{2}}}=2019\cdot\dfrac{2}{\sqrt{\dfrac{2019}{2}}}=2\sqrt{2}\sqrt{2019}\)

\(\Rightarrow P\ge\sqrt{2}\sqrt{2019}\)

Dấu = khi \(x=y=\dfrac{2019}{2}\)

1001 x 2019 - 2019 

= 2019 x ( 1001 -1 )

= 2019 x 1000

= 2019000

\(1001\cdot2019-2019=1001\cdot2019-2019\cdot1\)

\(=\left(1001-1\right)\cdot2019=1000\cdot2019\)

\(=2019000\)

k mk nha. 

#mon

2019000