Lời giải:

$\frac{x^3}{8}=\frac{y^3}{64}=\frac{z^3}{216}$
$\Rightarrow (\frac{x}{2})^3=(\frac{y}{4})^3=(\frac{z}{6})^3$
$\Rightarrow \frac{x}{2}=\frac{y}{4}=\frac{z}{6}$

$\Rightarrow \frac{x^2}{4}=\frac{y^2}{16}=\frac{z^2}{36}$

Áp dụng TCDTSBN:

$\frac{x^2}{4}=\frac{y^2}{16}=\frac{z^2}{36}=\frac{x^2+y^2+z^2}{4+16+36}=\frac{14}{56}=\frac{1}{4}$

$\Rightarrow x^2=1\Rightarrow x=\pm 1$

Nếu $x=1$ thì $\frac{y}{4}=\frac{z}{6}=\frac{1}{2}\Rightarrow y=2; z=3$

$\Rightarrow x+y-z=1+2-3=0$

Nếu $x=-1$ thì $\frac{y}{4}=\frac{z}{6}=\frac{-1}{2}\Rightarrow y=-2; z=-3$

$\Rightarrow x+y-z=(-1)+(-2)-(-3)=0$

Vậy $x+y-z=0$

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

x - 1/2 = y - 2/3 = z-3/4 = 2x - 2 + 3y - 6 - z + 3/4 + 9 - 4 = 95 + -5/10 = 10

x-1/2 = 10 => x =21

y-2/3  =10 => y = 32

z-3/4 = 10 => z = 43

Vậy x + y + z = 21 + 32 + 43 = 96

Ta có:

\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\frac{2\left(x-1\right)}{4}=\frac{3\left(y-2\right)}{9}=\frac{z-3}{4}\)

\(\Rightarrow\frac{2x-2.1}{4}=\frac{3y-3.2}{9}=\frac{z-3}{4}\)

\(\Rightarrow\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}\)

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

\(\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}=\frac{\left(2x-2\right)+\left(3y-6\right)-\left(z-3\right)}{4+9-4}=\frac{2x-2+3y-6-z+3}{9}=\frac{\left(2x+3y-z\right)+\left(-2+-6+3\right)}{9}=\frac{50+\left(-5\right)}{9}=\frac{45}{9}=5\)\(\Rightarrow\frac{x-1}{2}=5\Rightarrow x=5.2+1=11\)

\(\Rightarrow\frac{y-2}{3}=5\Rightarrow y=5.3+2=17\)

\(\Rightarrow\frac{z-3}{4}=5\Rightarrow z=5.4+3=23\)

Vậy \(x+y-z=11+17-23=28-23=5\)

Ta có: \(\frac{x-1}{2}=\frac{2x-2}{4};\frac{y-2}{3}=\frac{3y-6}{9}\) 

=> \(\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}\) và \(2x+3y-z=50\)

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

\(\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}=\frac{2x-2+3y-6-z+3}{4+9-4}\)

\(=\frac{2x+3y-z-\left(2+6-3\right)}{9}=\frac{50-5}{9}=5\)

=> \(x=5.2+1=11\)

\(y=5.3+2=17\)

\(z=5.4+3=23\)

\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\Rightarrow\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}.\)

Áp dụng tc dãy tỉ số bằng nhau ta có : 

\(\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}=\frac{2x-2+3y-6-z+3}{4+9-4}=\frac{2x+3y-z-5}{9}\)

\(=\frac{50-5}{9}=5\)

\(\left(+\right)\frac{x-1}{2}=5=>x=11\)

\(\left(+\right)\frac{y-2}{3}=5=>y=17\)

\(\left(+\right)\frac{z-3}{4}=5\Rightarrow z=23\)

\(=>x+y+z=11+17+23=51\)

\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\Rightarrow\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}\)

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

\(\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}=\frac{2x-2+3y-6-z+3}{4+9-4}=\frac{\left(2x+3y-z\right)+\left(-2-6+3\right)}{9}\)\(=\frac{50-5}{9}=\frac{45}{9}=5\)

Khi đó:\(\frac{2x-2}{4}=5\Rightarrow2x-2=20\Rightarrow x=11;\frac{3y-6}{9}=5\Rightarrow3y-6=45\Rightarrow y=17;\)

\(\frac{z-3}{4}=5\Rightarrow z-3=20\Rightarrow23\)

Áp dụng t/c vủa dãy tỉ số bằng nhau ta có \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\frac{x-1-2\left(y-2\right)+3\left(z-3\right)}{2-2.3+3.4}=\frac{\left(x-2y+3z\right)-1+4-9}{8}=\frac{14-6}{8}=1\)

=> x - 1 = 2; y - 2 = 3; z - 3 = 4

=> x = 3; y = 5; z = 7

Vậy...

Ta có : \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\) và x-2y+3z=14

=> \(\frac{x-1}{2}=\frac{2y-4}{6}=\frac{3z-9}{12}\)và x-2y+3z=14

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

​\(\frac{x-1}{2}=\frac{2y-4}{6}=\frac{3z-9}{12}\)​\(=\frac{x-2y+3z-14}{20}=\frac{14-14}{20}=0\)

Từ \(\frac{x-1}{2}=0=>x-1=0=>x=1\)

      \(\frac{2y-4}{6}=0=>2y-4=0=>2y=4=>y=2\)

      \(\frac{3z-9}{12}=0=>3z-9=0=>3z=9=>z=3\)

​​

Đặt \(a=\sqrt{2x-3}\) ; \(b=\sqrt{y-2}\) ; \(c=\sqrt{3z-1}\) (\(a,b,c>0\))

Ta có : \(\frac{1}{a}+\frac{4}{b}+\frac{16}{c}+a+b+c=14\)

\(\Leftrightarrow\left(\sqrt{2x-3}+\frac{1}{\sqrt{2x-3}}-2\right)+\left(\sqrt{y-2}+\frac{4}{\sqrt{y-2}}-4\right)+\left(\sqrt{3z-1}+\frac{16}{\sqrt{3z-1}}-8\right)=0\)

\(\Leftrightarrow\left[\frac{\left(2x-3\right)-2\sqrt{2x-3}+1}{\sqrt{2x-3}}\right]+\left[\frac{\left(y-2\right)-4\sqrt{y-2}+4}{\sqrt{y-2}}\right]+\left[\frac{\left(3z-1\right)-8\sqrt{3z-1}+16}{\sqrt{3z-1}}\right]=0\)

\(\Leftrightarrow\frac{\left(\sqrt{2x-3}-1\right)^2}{\sqrt{2x-3}}+\frac{\left(\sqrt{y-2}-2\right)^2}{\sqrt{y-2}}+\frac{\left(\sqrt{3z-1}-4\right)^2}{\sqrt{3z-1}}=0\)

\(\Leftrightarrow\hept{\begin{cases}\left(\sqrt{2x-3}-1\right)^2=0\\\left(\sqrt{y-2}-2\right)^2=0\\\left(\sqrt{3z-1}-4\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\y=6\\z=\frac{17}{3}\end{cases}}}\)(TMĐK)

Vậy : \(\left(x;y;z\right)=\left(2;6;\frac{17}{3}\right)\)

Phần đặt ẩn a,b,c bạn bỏ đi nhé ^^