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áp dụng t/c dãy tỉ số = nhau ta có:\
\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\frac{\left(x-1\right)-2\left(y-2\right)+3\left(z-3\right)}{2-2.3+3.4}=\frac{\left(x-2y+3z\right)-1+4-9}{8}=\frac{-10-6}{8}=-2\)
=>x-1=-4;y-2=-6;z-3=-8
=>x=-3;y=-4;z=-5
=>x+y+z=-12
Ta có: \(\frac{x-1}{2}=\frac{2x-2}{4};\frac{y-2}{3}=\frac{3y-6}{9};\frac{z-3}{4}=\frac{-z+3}{-4}\)
Vì \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\)nên \(\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{-z+3}{-4}=\frac{2x-2+3y-6-z+3}{4+9+\left(-4\right)}=\frac{50-5}{9}=\frac{45}{9}=5\)
\(\Rightarrow\frac{x-1}{2}=5\Rightarrow x=11\)
\(\Rightarrow\frac{y-2}{3}=5\Rightarrow y=17\)
\(\Rightarrow\frac{z-3}{4}=5\Rightarrow z=23\)
Vậy, \(x+y+z=11+17+23=51\)
\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\frac{2\left(x-1\right)+3\left(y-2\right)-\left(z-3\right)}{2.2+3.3-4}=\frac{95-5}{9}=10\)
x-1 =20 => x =21
y-2 =30 => y =32
z-3 =40 => z =43
\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\frac{2x-2+3y-6-z+3}{4+9-4}=\frac{90}{9}=10\)
\(\Rightarrow\) x - 1 = 20; y - 2 = 30; z - 3 = 40
\(\Rightarrow\) x = 21; y = 32; z = 43
\(\frac{x^3}{1000}=\frac{y^3}{3375}=\frac{z^3}{1728}=>\left(\frac{x}{10}\right)^3=\left(\frac{y}{15}\right)^3=\left(\frac{z}{12}\right)^3=>\frac{x}{10}=\frac{y}{15}=\frac{z}{12}=\frac{x-y+z}{10-15+12}=-\frac{49}{7}=-7\)
=>x=-70;y=-105;z=-84
=>x+y+z=-259
tick tớ nhé