Bài 1.

Ta có:\(\left(x+\sqrt{x^2+2020}\right)\left(\sqrt{x^2+2020}-x\right)=x^2+2020-x^2=2020\)

\(\Rightarrow\left(x+\sqrt{x^2+2020}\right)\left(y+\sqrt{y^2+2020}\right)=\left(x+\sqrt{x^2+2020}\right)\left(\sqrt{x^2+2020}-x\right)\)

\(\Rightarrow y+\sqrt{y^2+2020}=\sqrt{x^2+2020}-x\)

\(\Rightarrow x+y=\sqrt{x^2+2020}-\sqrt{y^2+2020}\)   (1)

Ta có:\(\left(y+\sqrt{y^2+2020}\right)\left(\sqrt{y^2+2020}-y\right)=y^2+2020-y^2=2020\)

\(\Rightarrow\left(x+\sqrt{x^2+2020}\right)\left(y+\sqrt{y^2+2020}\right)=\left(y+\sqrt{y^2+2020}\right)\left(\sqrt{y^2+2020}-y\right)\)

\(\Rightarrow x+\sqrt{x^2+2020}=\sqrt{y^2+2020}-y\)

\(\Rightarrow x+y=\sqrt{y^2+2020}-\sqrt{x^2+2020}\)          (2)

Cộng vế với vế của (1) và (2) ta có:

\(2\left(x+y\right)=\sqrt{y^2+2020}-\sqrt{x^2+2020}+\sqrt{x^2+2020}-\sqrt{y^2+2020}\)

\(\Rightarrow2\left(x+y\right)=0\Rightarrow x+y=0\)

Bài 2: 

Ta có: (2a+1)(2b+1)=9

nên \(2b+1=\dfrac{9}{2a+1}\)

\(\Leftrightarrow2b=\dfrac{9}{2a+1}-\dfrac{2a+1}{2a+1}=\dfrac{8-2a}{2a+1}\)

\(\Leftrightarrow b=\dfrac{8-2a}{4a+2}=\dfrac{4-a}{2a+1}\)

\(\Leftrightarrow b+2=\dfrac{4-a+4a+2}{2a+1}=\dfrac{3a+6}{2a+1}\)

Ta có: \(A=\dfrac{1}{a+2}+\dfrac{1}{b+2}\)

\(=\dfrac{1}{a+2}+\dfrac{2a+1}{3a+6}\)

\(=\dfrac{3+2a+1}{3a+6}\)

\(=\dfrac{2a+4}{3a+6}=\dfrac{2}{3}\)

a) Quy luật là gì ??

b) 

Đặt

 \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2020}}\\\Rightarrow2A=1+\dfrac{1}{2}+...+\dfrac{1}{2^{2019}}\\ \Rightarrow2A-A=1-\dfrac{1}{2^{2020}}\Rightarrow A=1-\dfrac{1}{2^{2020}}\)

Suy ra , phương trình trở thành :

213 -x  =13

<=> x=200

\(y=\dfrac{1}{2x^2+x-1}=\dfrac{1}{\left(x+1\right)\left(2x-1\right)}=\dfrac{2}{3}.\dfrac{1}{2x-1}-\dfrac{1}{3}.\dfrac{1}{x+1}\)

\(y'=\dfrac{2}{3}.\dfrac{-2}{\left(2x-1\right)^2}-\dfrac{1}{3}.\dfrac{-1}{\left(x+1\right)^2}=\dfrac{2}{3}.\dfrac{\left(-1\right)^1.2^1.1!}{\left(2x-1\right)^2}-\dfrac{1}{3}.\dfrac{\left(-1\right)^1.1!}{\left(x+1\right)^2}\)

\(y''=\dfrac{2}{3}.\dfrac{\left(-1\right)^2.2^2.2!}{\left(2x-1\right)^3}-\dfrac{1}{3}.\dfrac{\left(-1\right)^2.2!}{\left(x+1\right)^3}\)

\(\Rightarrow y^{\left(n\right)}=\dfrac{2}{3}.\dfrac{\left(-1\right)^n.2^n.n!}{\left(2x-1\right)^{n+1}}-\dfrac{1}{3}.\dfrac{\left(-1\right)^n.n!}{\left(x+1\right)^{n+1}}\)

\(\Rightarrow y^{\left(2019\right)}=\dfrac{2}{3}.\dfrac{\left(-1\right)^{2019}.2^{2019}.2019!}{\left(2x-1\right)^{2020}}-\dfrac{1}{3}.\dfrac{\left(-1\right)^{2019}.2019!}{\left(x+1\right)^{2020}}\)

\(=\dfrac{2019!}{3}\left(\dfrac{1}{\left(x+1\right)^{2020}}-\dfrac{2^{2020}}{\left(2x-1\right)^{2020}}\right)\)

\(y=\dfrac{1}{3x^2-x-2}=\dfrac{1}{\left(x-1\right)\left(3x+2\right)}=\dfrac{1}{5}.\dfrac{1}{x-1}-\dfrac{3}{5}.\dfrac{1}{3x+2}\)

\(y'=\dfrac{1}{5}.\dfrac{\left(-1\right)^1.1!}{\left(x-1\right)^2}-\dfrac{3}{5}.\dfrac{\left(-1\right)^1.3^1.1!}{\left(3x+2\right)^2}\)

\(y''=\dfrac{1}{5}.\dfrac{\left(-1\right)^2.2!}{\left(x-1\right)^3}-\dfrac{3}{5}.\dfrac{\left(-1\right)^2.3^2.2!}{\left(3x+2\right)^3}\)

\(\Rightarrow y^{\left(n\right)}=\dfrac{1}{5}.\dfrac{\left(-1\right)^n.n!}{\left(x-1\right)^{n+1}}-\dfrac{3}{5}.\dfrac{\left(-1\right)^n.3^n.n!}{\left(3x+2\right)^{n+1}}\)

\(\Rightarrow y^{\left(2019\right)}=\dfrac{1}{5}.\dfrac{\left(-1\right)^{2019}.2019!}{\left(x-1\right)^{2020}}-\dfrac{3}{5}.\dfrac{\left(-1\right)^{2019}.3^{2019}.2019!}{\left(3x+2\right)^{2019}}\)

\(=\dfrac{2019!}{5}\left(\dfrac{3^{2020}}{\left(3x+2\right)^{2020}}-\dfrac{1}{\left(x-1\right)^{2020}}\right)\)

a)

`(2x-1)(x+2/3)=0`

\(< =>\left[{}\begin{matrix}2x-1=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{2}{3}\end{matrix}\right.\)

b)

\(\dfrac{x+4}{2019}+\dfrac{x+3}{2020}=\dfrac{x+2}{2021}+\dfrac{x+1}{2022}\)

\(< =>\dfrac{x+4}{2019}+1+\dfrac{x+3}{2020}+1=\dfrac{x+2}{2021}+1+\dfrac{x+1}{2022}+1\)

\(< =>\dfrac{x+2023}{2019}+\dfrac{x+2023}{2020}=\dfrac{x+2023}{2021}+\dfrac{x+2023}{2022}\)

\(< =>\left(x+2023\right)\left(\dfrac{1}{2019}+\dfrac{1}{2020}-\dfrac{1}{2021}-\dfrac{1}{2022}\right)=0\)

\(< =>x+2023=0\left(\dfrac{1}{2019}+\dfrac{1}{2020}-\dfrac{1}{2021}-\dfrac{1}{2022}\ne0\right)\\ < =>x=-2023\)

sai rồi , x không thể có 2 giá trị

a) + Chia thành 2 trường hợp 

- 2x - 1 = 0

2x = 0 + 1

2x = 1

x = 1 : 2

x = 0,5

- x + 2/3 = 0

x = 0 - 2/3

x = -2/3

vậy x = { 0,5 ; -2/3 }

Cứu với ;-;

Bài 2: 

Ta có: \(11^{1979}< 11^{1980}=1331^{660}\)

\(37^{1320}=37^{2\cdot660}=1369^{660}\)

mà \(1331^{660}< 1369^{660}\)

nên \(11^{1979}< 37^{1320}\)

\(ĐKXĐ:\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

\(P\left(x\right)=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)

\(P\left(x\right)=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)

\(P\left(x\right)=x-\sqrt{x}-2\sqrt{x}-2+2\sqrt{x}+2\)

\(P\left(x\right)=x-\sqrt{x}\)

Ta có : \(\dfrac{P\left(x\right)}{2020\sqrt{x}}=\dfrac{x-\sqrt{x}}{2020\sqrt{x}}=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{2020\sqrt{x}}=\dfrac{\sqrt{x}-1}{2020}\)

Để \(\dfrac{P\left(x\right)}{2020\sqrt{x}}min\Leftrightarrow\dfrac{\sqrt{x}-1}{2020}min\Leftrightarrow\sqrt{x}-1\) min (vì 2020 > 0)

Lại có : \(\sqrt{x}-1\ge-1\forall x\)

Dấu "=" xảy ra <=> x = 0

Vậy Min\(\dfrac{P\left(x\right)}{2020\sqrt{x}}=\dfrac{-1}{2020}\Leftrightarrow x=0\)