Cứu với ;-;

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 }

\(A=\left(\dfrac{2020}{2021}xy^5z\right).\left(\dfrac{2020}{2021}x^3yz^2\right).\left(-\dfrac{2020}{2021}\right)^0\)

\(a)A=\dfrac{2020.2021.2020}{2021.2020.2021}.\left(x.x^3\right).\left(y^5.y\right).\left(z.z^2\right)\Leftrightarrow A=\dfrac{2020}{2021}x^4.y^6.z^3\)

\(b)A=\dfrac{2020}{2021}x^4.y^6.z^3\)

\(\Rightarrow\text{A có hệ số là:}\dfrac{2020}{2021}\)

\(\text{Phần biến là:}\left(x,y,z\right)\)

\(c)\text{Xét A ta có:}\dfrac{2020}{2021}< 0;x^4,y^6\text{ luôn }< 0\)

\(\Rightarrow\dfrac{2020}{2021}x^4.y^6>0\Rightarrow\text{ Nếu }z< 0\Rightarrow A\le0\text{ và z có số mũ là:3}\)

\(\text{Chẳng hạn:}\left(-\right).\left(-\right).\left(-\right)=\left(-\right).< 0\Rightarrow z\text{ phải }\ge0\text{ thì }A\ge0\)

\(\Rightarrow Z\in N\)

đề bài yêu cầu gì bn? 

TH1: x<2020

Q=2020-x+2021-x+2022-x=6063-3x

TH2: 2020<=x<2021

Q=x-2020+2021-x+2022-x=2023-x

TH3: 2021<=x<2022

Q=x-2020+x-2021+2022-x=x-2019

TH4: x>=2022

Q=x-2020+x-2021+x-2022=3x-6063

\(S=\dfrac{x^3}{16\left(y+16\right)}+\dfrac{y^3}{16\left(x+16\right)}+\dfrac{2021}{2022}\)

\(\dfrac{x^3}{16\left(y+16\right)}+\dfrac{y+16}{100}+\dfrac{16}{80}\ge3\sqrt[3]{\dfrac{x^3\left(y+16\right).16}{16\left(y+16\right).100.80}}=\dfrac{3x}{20}\)

\(tương\) \(tự\Rightarrow\dfrac{y^3}{16\left(x+16\right)}\ge\dfrac{3y}{20}\)

\(\Rightarrow S\ge\dfrac{3x}{20}+\dfrac{3y}{20}-\left(\dfrac{x+16}{100}+\dfrac{y+16}{100}\right)-2.\dfrac{16}{80}+\dfrac{2021}{2022}=\dfrac{3x+3y}{20}-\dfrac{x+y+32}{100}-\dfrac{2}{5}+\dfrac{2021}{2022}=\dfrac{15x+15y-x-y-32}{100}-\dfrac{2}{5}+\dfrac{2021}{2022}=\dfrac{14\left(x+y\right)-32}{100}-\dfrac{2}{5}+\dfrac{2021}{2022}\)

\(xy=16\le\dfrac{\left(x+y\right)^2}{4}\Rightarrow x+y\ge8\Rightarrow S\ge\dfrac{14.8-32}{100}-\dfrac{2}{5}+\dfrac{2021}{2022}=\dfrac{2}{5}+\dfrac{2021}{2022}\)

\(\Rightarrow minS=\dfrac{2}{5}+\dfrac{2021}{2022}\Leftrightarrow x=y=4\)

\(\dfrac{x^3}{16\left(y+16\right)}+\dfrac{y+16}{100}+\dfrac{1}{5}\ge3\sqrt[3]{\dfrac{x^3\left(y+16\right)}{16.100.5\left(y+16\right)}}=\dfrac{3x}{20}\)

Tương tự: \(\dfrac{y^3}{16\left(x+16\right)}+\dfrac{x+16}{100}+\dfrac{1}{5}\ge\dfrac{3y}{20}\)

Cộng vế:

\(S+\dfrac{x+y+32}{100}+\dfrac{2}{5}\ge\dfrac{3\left(x+y\right)}{20}+\dfrac{2021}{2022}\)

\(S\ge\dfrac{9}{20}\left(x+y\right)-\dfrac{42}{25}+\dfrac{2021}{2022}\ge\dfrac{9}{20}.2\sqrt{xy}-\dfrac{42}{25}+\dfrac{2021}{2022}=...\)

\(=\dfrac{tan\left(\dfrac{pi}{2}+x\right)\cdot sin\left(-x\right)\cdot cos\left(x-pi\right)}{cos\left(\dfrac{pi}{2}-x\right)\cdot sin\left(x+pi\right)}\)

\(=\dfrac{-cotx\cdot sin\left(-x\right)\cdot\left(-cosx\right)}{sinx\cdot-sinx}\)

\(=\dfrac{cotx\cdot sinx\left(-1\right)\cdot cosx}{-sinx\cdot sinx}=\dfrac{\dfrac{cosx}{sinx}\cdot cosx}{sinx}=\dfrac{cos^2x}{sin^2x}=cot^2x\)