Ta có; \(y^2=x\cdot z\)

=>\(\frac{x}{y}=\frac{y}{z}\) (1)

\(z^2=y\cdot t\)

=>\(\frac{y}{z}=\frac{z}{t}\) (2)

Từ (1),(2) suy ra \(\frac{x}{y}=\frac{y}{z}=\frac{z}{t}=k\)

=>\(\begin{cases}z=tk\\ y=zk=tk\cdot k=tk^2\\ x=y\cdot k=tk^2\cdot k=tk^3\end{cases}\)

\(\frac{x^3+y^3-2z^3}{y^3+z^3-2t^3}=\frac{\left(yk\right)^3+\left(zk\right)^3-2\cdot\left(tk\right)^3}{y^3+z^3-2t^3}=\frac{k^3\left(y^3+z^3-2t^3\right)}{y^3+z^3-2t^3}=k^3\)

\(\left(\frac{x+y-2z}{y+z-2t}\right)^3=\left(\frac{yk\cdot zk-2\cdot tk}{y+z-2t}\right)^3=k^3\)

Do đó: \(\frac{x^3+y^3-2z^3}{y^3+z^3-2t^3}=\left(\frac{x+y-2z}{y+z-2t}\right)^3\)

sao ko ai trả lời vậy

a) Ta có:

\(\dfrac{x+1}{3}=\dfrac{y+2}{2}=\dfrac{z+3}{1}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta được

\(\dfrac{x+1}{3}=\dfrac{y+2}{2}=\dfrac{z+3}{1}\)

\(=\dfrac{x+1-y-2+z+3}{3-2+1}\)

\(=\dfrac{22+2}{2}\)

\(=\dfrac{24}{2}=12\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x+1}{3}=12\\\dfrac{y+2}{2}=12\\\dfrac{z+3}{1}=12\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x+1=36\\y+2=24\\z+3=12\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=36-1=35\\y=24-2=22\\z=12-3=9\end{matrix}\right.\)

BIK CHẾT LIỀN