\(\frac{x^3}{1000}=\frac{y^3}{3375}=\frac{z^3}{1728}\Rightarrow\left(\frac{x}{10}\right)^3=\left(\frac{y}{15}\right)^3=\left(\frac{z}{12}\right)^3\)

\(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{12}=\frac{x-y+z}{10-15+12}\left(\text{t/c dãy tỉ số = nhau}\right)=\frac{-49}{7}=-7\)

\(\Rightarrow\frac{x}{10}=-7\Rightarrow x=-7.10=-70\)

\(\Rightarrow\frac{y}{15}=-7\Rightarrow y=-7.15=-105\)

\(\Rightarrow\frac{z}{12}=-7\Rightarrow z=-7.12=-84\)

Vậy x+y+z=(-70)+(-105)+(-84)=-259.

\(\frac{x^3}{1000}=\frac{y^3}{3375}=\frac{z^3}{1728}\)

\(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{12}=\frac{x-y+z}{10-15+12}=\frac{-49}{7}=-7\)

\(\Rightarrow\) x = - 70; y = - 105; z = - 84

\(\Rightarrow\) x + y + z = - 259

\(Q=\frac{x^3}{\left(x-y\right)\left(x-z\right)}+\frac{y^3}{\left(y-x\right)\left(y-z\right)}+\frac{z^3}{\left(z-x\right)\left(z-y\right)}\)

\(=\frac{x^3}{\left(x-y\right)\left(x-z\right)}-\frac{y^3}{\left(x-y\right)\left(y-z\right)}+\frac{z^3}{\left(x-z\right)\left(y-z\right)}\)

\(=\frac{x^3\left(y-z\right)-y^3\left(x-z\right)+z^3\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)(1)

Ta có: 

      \(x^3\left(y-z\right)-y^3\left(x-z\right)+z^3\left(x-y\right)\)

\(=x^3\left(y-z\right)-y^3\left(y-z\right)-y^3\left(x-y\right)+z^3\left(x-y\right)\)

\(=\left(y-z\right)\left(x^3-y^3\right)-\left(x-y\right)\left(y^3-z^3\right)\)

\(=\left(y-z\right)\left(x-y\right)\left(x^2+xy+y^2\right)-\left(x-y\right)\left(y-z\right)\left(y^2+yz+z^2\right)\)

\(=\left(x-y\right)\left(y-z\right)\left(x^2+xy+y^2-y^2-yz-z^2\right)\)

\(=\left(x-y\right)\left(y-z\right)\left(x^2+xy-yz-z^2\right)\)

\(=\left(x-y\right)\left(y-z\right)\left[\left(x-z\right)\left(x+z\right)+y\left(x-z\right)\right]\)

\(=\left(x-y\right)\left(y-z\right)\left(x-z\right)\left(x+y+z\right)=1000\left(x-y\right)\left(y-z\right)\left(x-z\right)\)(2)

Từ (1) và (2), ta có Q = 1000

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

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

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

bạn tách phần tử số ra 2 phần: xong thì cộng theo giả thiết là phần ẩn rồi cộng thêm phần thứ 2

tíc mình nha

Ta có:\(\frac{x-1}{2}=\frac{2.\left(x-1\right)}{2.2}=\frac{2x-2}{4}\)

\(\frac{y-2}{3}=\frac{3.\left(y-2\right)}{3.3}=\frac{3y-6}{9}\)

Theo t/c dãy tỉ số = nhau:

\(\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}=\frac{2x-2+3y-6-z+3}{4+9-4}=\frac{2x+3y-z-5}{9}=\frac{95-5}{9}=\frac{90}{9}=10\)

=> \(\frac{x-1}{2}=10\Rightarrow x-1=10.2=20\Rightarrow x=20+1=21\)

=> \(\frac{y-2}{3}=10\Rightarrow y-2=10.3=30\Rightarrow y=30+2=32\)

=> \(\frac{z-3}{4}=10\Rightarrow z-3=10.4=40\Rightarrow z=40+3=43\)

Vậy x + y + z = 21 + 32 + 43 = 96.