\(\frac{12-3x}{32}=\frac{6}{4-x}\)

\(\Rightarrow\left(12-3x\right)\left(4-x\right)=6\times32\)

\(\Rightarrow3\left(4-x\right)\left(4-x\right)=192\)

\(\Rightarrow3\left(4-x\right)^2=192\)

\(\Rightarrow\left(4-x\right)^2=192\div3\)

\(\Rightarrow\left(4-x\right)^2=64\)

\(\Rightarrow\left(4-x\right)^2=8^2\)

\(\Rightarrow\orbr{\begin{cases}4-x=-8\\4-x=8\end{cases}\Rightarrow\orbr{\begin{cases}x=4+8\\x=4-8\end{cases}\Rightarrow}\orbr{\begin{cases}x=12\\x=-4\end{cases}}}\)

    Vậy \(x=12;x=-4\)

Ta có:  |x - 2018| + |2019 - x| ≥ |x - 2018 + 2019 - x| = |1| = 1

Dấu " = " xảy ra <=> (x - 2018)(2019 - x) ≥ 0

Th1: \(\hept{\begin{cases}x-2018\text{ }\ge0\\2019-x\ge0\end{cases}\Rightarrow}\hept{\begin{cases}x\ge2018\\x\le2019\end{cases}}\)

Th2: \(\hept{\begin{cases}x-2018\text{ }\le0\\2019-x\le0\end{cases}\Rightarrow}\hept{\begin{cases}x\le2018\\x\ge2019\end{cases}}\)(Vô lý)

Vậy GTNN  |x - 2018| + |2019 - x| = 1 khi 2018 ≤ x ≤ 2019

THANKYOU

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

\(\Rightarrow-3\cdot\left(2-x\right)=4\cdot\left(3x-1\right)\)

\(\Rightarrow-6+3x=12x-4\)

\(\Rightarrow3x-12x=-4+6\)

\(\Rightarrow-9x=2\)

\(\Rightarrow x=-\frac{2}{9}\)

*

giải

Ta có: \(\frac{2-x}{4}=\frac{3x-1}{-3}\)

=>\(\left(2-x\right)\cdot\left(-3\right)=\left(3x-1\right)\cdot4\)

=>\(-6+3x=12x-4\)

=>\(-6+4=12x-3x\)

=>\(-2=9x\)

=> \(x=-\frac{2}{9}\)

#CHÚC BẠN HỌC TỐT

=>81.(x+3)=125.729

=>81.(x+3)=91125

=>x+3=91125:81

=>x+3=1125

=>x=1125:3

=>x=375

vậy x=375