\(4x^3-3=29\Rightarrow4x^3=32\Rightarrow x^3=8\Rightarrow x=2\)
Thay vào: \(\frac{x+16}{9}=\frac{y-25}{-16}=\frac{z+49}{25}\)
\(\Rightarrow\frac{2+16}{9}=\frac{y-25}{-16}=\frac{z+49}{25}\Rightarrow\frac{y-25}{-16}=\frac{z+49}{25}=2\)
\(\Rightarrow\left\{{}\begin{matrix}y=2.\left(-16\right)+25=-7\\z=2.25-49=1\end{matrix}\right.\)
\(\Rightarrow x-2y+3z=2-2.\left(-7\right)+3.1=2+14+3=19\)
\(4x^3-3=29\)
\(\Rightarrow4x^3=32\Rightarrow x^3=8\Rightarrow x=2\)
\(\Rightarrow\frac{x+16}{9}=\frac{2+16}{9}=2\)
Ta có
\(\frac{x+16}{9}=\frac{y-25}{-16}\Rightarrow\frac{y-25}{-16}=2\)
\(\Rightarrow y-25=-32\Rightarrow y=-7\);
\(\frac{x+16}{9}=\frac{z+49}{25}\Rightarrow\frac{z+49}{25}=2\Rightarrow z+49=50\Rightarrow z=1\)
Vậy x = 2 ; y = -7 ; z = 1
\(\dfrac{x+16}{9}=\dfrac{y-25}{-16}=\dfrac{z+49}{25}\) (1)
Ta có: \(4x^3-3=29\)
\(\Rightarrow4x^3=32\Rightarrow x^3=8\)
\(\Rightarrow x=2\)
Thay \(x=2\) vào điều (1) ta có:
\(\dfrac{2+16}{9}=\dfrac{y-25}{-16}=\dfrac{z+49}{25}\)
\(\Rightarrow\dfrac{y-25}{-16}=\dfrac{z+49}{25}=\dfrac{18}{9}\)
\(\Rightarrow\dfrac{y-25}{-16}=\dfrac{z+49}{25}=2\)
\(\Rightarrow\left\{{}\begin{matrix}y-25=2.\left(-16\right)\\z+49=2.25\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y-25=-32\\z+49=50\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}y=-7\\z=1\end{matrix}\right.\)
Vậy giá trị của biểu thức \(x-2y+3z\) là:
\(2-2.\left(-7\right)+3.1=2+14+3=19\)
