\(A=\frac{\left|x-2019\right|+2020-2}{\left|x-2019\right|+2020}=1-\frac{2}{\left|x-2019\right|+2020}\)

Vì \(\left|x-2019\right|\ge0\)

=> \(\left|x-2019\right|+2020\ge2020\)

=> \(\frac{2}{\left|x-2019\right|+2020}\le\frac{2}{2020}\)

=> \(-\frac{2}{\left|x-2019\right|+2020}\ge-\frac{2}{2020}\)

=> \(1-\frac{2}{\left|x-2019\right|+2020}\ge1-\frac{2}{2020}=\frac{2018}{2020}=\frac{1009}{1010}\)

=> \(A\ge\frac{1009}{1010}\)

Dấu "=" xảy ra <=> \(x-2019=0\Leftrightarrow x=2019\)

Vậy GTNN của A bằng 1009/1010 đạt tại x = 2019.

\(X-\frac{3}{2}\times\left(-1\right)=\frac{4}{3}\)

\(X-\frac{3}{2}=\frac{4}{3}\div\left(-1\right)\)

\(X-\frac{3}{2}=\frac{4}{3}\div\left(-\frac{3}{3}\right)\)

\(X-\frac{3}{2}=\frac{4}{3}\times\left(-\frac{3}{3}\right)\)

\(X-\frac{3}{2}=-\frac{12}{9}=-\frac{4}{3}\)

\(X=-\frac{4}{3}+\frac{3}{2}\)

\(X=-\frac{8}{6}+\frac{9}{6}\)

\(X=\frac{-8+9}{6}=\frac{1}{6}\)

(sai thì thôi)

\(x-\frac{3}{2}.\left(-1\right)=\frac{4}{3}\)

\(x-\frac{3}{2}=\frac{4}{3}.\left(-1\right)\)

\(x-\frac{3}{2}=-\frac{4}{3}\)

\(x=-\frac{4}{3}+\frac{3}{2}\)

\(x=-\frac{8}{6}+\frac{9}{6}\)

\(x=\frac{1}{6}\)

\(\left|3x-0,\left(36\right)\right|+\left|5y+1,2\left(18\right)\right|=0\)

Có: \(\left|3x-0,\left(36\right)\right|\ge0,\forall x\)

       \(\left|5y+1,2\left(18\right)\right|\ge0,\forall y\)

=>\(\left|3x-0,\left(36\right)\right|+\left|5y+1,2\left(18\right)\right|\ge0;\forall x,y\)

Dấu "=" xảy ra: \(\hept{\begin{cases}3x-0,\left(36\right)=0\\5y+1,2\left(18\right)=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=0,\left(12\right)\\y=0,24\left(36\right)\end{cases}}\)

a)Ta có:

\(1-\frac{2008}{2009}=\frac{1}{2009},1-\frac{2018}{2019}=\frac{1}{2019}\)

Từ đó ta suy ra:\(\frac{2008}{2009}< \frac{2018}{2019}\)

vì \(\frac{1}{2009}>\frac{1}{2019}\)

Ta có:

\(\frac{12x-15y}{7}=\frac{20z-12x}{9}=\frac{15y-20z}{11}=\frac{12x-15y+20z-12x+15y-20z}{7+9+11}=0\)

Đề bài gì lạ vậy

Áp dụng tính chất của dãy tỉ số bằng nhau ta có : 

\(\frac{12x-15y}{7}=\frac{20z-12x}{9}=\frac{15y-20z}{11}=\frac{12x-15y+20z-12x+15y-20z}{7+9+11}=\frac{0}{27}=0\)

\(\Rightarrow\hept{\begin{cases}12x-15y=0\\15y-20z=0\end{cases}\Rightarrow\hept{\begin{cases}12x=15y\\15y=20z\end{cases}\Rightarrow}\hept{\begin{cases}\frac{x}{15}=\frac{y}{12}\\\frac{y}{20}=\frac{z}{15}\end{cases}\Rightarrow}\hept{\begin{cases}\frac{x}{75}=\frac{y}{60}\\\frac{y}{60}=\frac{z}{45}\end{cases}\Rightarrow}\frac{x}{75}=\frac{y}{60}=\frac{z}{45}}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có : 

\(\frac{x}{75}=\frac{y}{60}=\frac{z}{45}=\frac{x+y+z}{75+60+45}=\frac{48}{180}=\frac{4}{15}\)

=> x = 75.4 : 15 = 20 ; 

     y = 60.4 : 15 = 16

     z = 45.4 : 15 = 12

Vậy x = 20 ; y = 16 ; z = 12