ko biết

ai bt giúp mik cái

x²⁰¹⁹-2019.x²⁰¹⁸+2019.x²⁰¹⁸+2019.x²⁰¹⁷-2019.x²⁰¹⁶+......2019.x-200     Tại x=2018

Giúp mik vs nhé 

Sai đề nên t sửa luôn nhé!

Vì \(x=2018\Rightarrow2019=2018+1=x+1\)

\(A=x^{2017}-2019\cdot x^{2018}+2019\cdot x^{2017}-2019\cdot x^{2016}+....+2019\cdot x-200\)

\(\Rightarrow A=x^{2019}-\left(x+1\right)x^{2018}+\left(x+1\right)x^{2017}-\left(x+1\right)x^{2016}+....-\left(x+1\right)x^2+\left(x+1\right)x-200\)

\(\Rightarrow A=x^{2019}-x^{2019}-x^{2018}+x^{2018}+x^{2017}-x^{2017}-x^{2016}+....-x^3-x^2+x^2+x-200\)

\(\Rightarrow A=x-200=2018-200=1818\)

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

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

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

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

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: 21^15=3^15*7^15

27^5*49^8=3^15*7^14

mà 15>14

nên 21^15>27^5*49^8

b: \(2020^{2020}-2020^{2019}=2020^{2019}\left(2020-1\right)=2020^{2019}\cdot2019\)

\(2020^{2019}-2020^{2018}=2020^{2018}\cdot2019\)

mà 2019>2018

nên 2020^2020-2020^2019>2020^2019-2020^2018

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

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

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

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

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 \(x=y=\dfrac{2019}{2}\)

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)

Vì \(x=2018\Rightarrow x+1=2019\)

Thay x+1=2019 vào biểu thức A  ta được :

\(A=x^6-\left(x+1\right)x^5+\left(x+1\right)x^4-...-\left(x+1\right)x+x+1\)

\(=x^6-x^6-x^5+x^5+x^4-...-x^2-x+x+1\)

\(=1\)

\(A=x^6-2019x^5+2018x^4-2019x^3+2019x^2-2019x+2019\)

\(=x^6-2018x^5-x^5+2018x^4+x^4-2018x^3-x^3+2018x^2+x^2\)

\(-2018x-x+2019\)

\(=x^5\left(x-2018\right)-x^4\left(x-2018\right)-x^3\left(x-2018\right)+x^2\left(x-2018\right)\)

\(+x\left(x-2018\right)-\left(x-2018\right)+1\)

= 1

Vì \(x=2018\Rightarrow x+1=2019\)

Thay \(x+1=2019\) vào biểu thức \(A\) ta được :

\(A=x^6-\left(x+1\right)x^5+\left(x+1\right)x^4-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+\left(x-1\right)\)

\(=x^6-x^6-x^5+x^5-x^4+x^4-x^3+x^3-x^2+x^2-x+x+1\)

\(=1\)

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

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

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

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

\(A=1-1\)

\(A=0.\)

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

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

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

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

\(A=1-1\)

\(A=0\)