\(\dfrac{a}{1}=\dfrac{b}{2}=\dfrac{c}{3}=\dfrac{4a-3b+2c}{4-6+6}=\dfrac{36}{4}=9\\ \Rightarrow\left\{{}\begin{matrix}a=9\\b=18\\c=27\end{matrix}\right.\\ \dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{5}=\dfrac{z}{4}\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{16}=\dfrac{x-y+z}{10-15+16}=\dfrac{-49}{11}\\ \Rightarrow\left\{{}\begin{matrix}x=-\dfrac{490}{11}\\y=-\dfrac{735}{11}\\z=-\dfrac{784}{11}\end{matrix}\right.\)

a)−3a⁵b²c² âm khi a sẽ dương.
3a²bc âm khi bc âm .
−2a³b5c=−2a³b⁴bc ta có a³ dương bcbc âm b⁴ dương ⇒−2a³b⁴bc dương .
⇒ Các số thõa mãn đề bài không thể cùng âm.
 

Đặt a/2=b/3=c/4=k

=>a=2k; b=3k; c=4k

Ta có: \(a^2+3b^2-2c^2=-16\)

\(\Leftrightarrow4k^2+27k^2-32k^2=-16\)

\(\Leftrightarrow k^2=16\)

Trường hợp 1: k=4

=>a=8; b=12; c=16

Trường hợp 2: k=-4

=>a=-8; b=-12; c=-16

REFER

\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\)

\(\Rightarrow\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{c^2}{16}=\dfrac{3b^2}{27}=\dfrac{2c^2}{32}=\dfrac{a^2+3b^2-2c^2}{4+27-32}=\dfrac{-16}{-1}=16\)

\(\Rightarrow\left\{{}\begin{matrix}a^2=64\\b^2=144\\c^2=256\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=\pm8\\b=\pm\\c=\pm16\end{matrix}\right.12}\)

Vậy (a; b; c)\(\in\){(8; 12; 16)}; {(-8; -12; -16)}

\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\Rightarrow\left\{{}\begin{matrix}b=\dfrac{3}{2}a\\c=2a\end{matrix}\right.\).

Ta có: \(a^2+3b^2-2c^2=a^2+3.\left(\dfrac{3}{2}a\right)^2-2.\left(2a\right)^2=-\dfrac{1}{4}a^2=-16\) \(\Rightarrow\) a=\(\pm\)8 \(\Rightarrow\) b=\(\pm\)12, c=\(\pm\)16.

a/

\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{2a}{3b}=\frac{2c}{3d}\Rightarrow\frac{2a}{2c}=\frac{3b}{3d}=\frac{2a+3b}{2c+3d}=\frac{2a-3b}{2c-3d}\)

\(\Rightarrow\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\left(dpcm\right)\)

b/

\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a^2}{ab}=\frac{c^2}{cd}\Rightarrow\frac{a^2}{c^2}=\frac{ab}{cd}\left(1\right)\)

\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{ab}{b^2}=\frac{cd}{d^2}\Rightarrow\frac{b^2}{d^2}=\frac{ab}{cd}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\frac{ab}{cd}=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2-b^2}{c^2-d^2}\left(dpcm\right)\)

Ta có: \(a^2+b^2+c^2+4\le ab+3b+2c\)

Hay: \(\left(a^2-ab+\frac{b^2}{4}\right)+\left(\frac{3b^2}{4}-3b+3\right)+\left(c^2-2c+1\right)\le0\)

\(\Leftrightarrow\left(a-\frac{b}{2}\right)^2+3\left(\frac{b}{2}-1\right)^2+\left(c-1\right)^2\le0\)

\(\Leftrightarrow\left\{{}\begin{matrix}c-1=0\\\frac{b}{2}-1=0\\a-\frac{b}{2}=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=2\\c=1\end{matrix}\right.\)

Vậy .....................