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3) 2x3-1=15 <=> x3=16/2=8=23 => x=2
\(\frac{x+16}{9}=\frac{y-25}{16}=\frac{z+9}{25}=\frac{x+16+y-25+z+9}{9+16+25}=\frac{x+y+z}{50}\)
=> \(\frac{x+16}{9}=\frac{x+y+z}{50}\)=> x+y+z=\(\frac{50\left(x+16\right)}{9}\)=\(\frac{50\left(2+16\right)}{9}=\frac{50.18}{9}=50.2=100\)
Vậy x+y+z=100
2x3 - 1 = 15 <=> 2x3 = 16
<=> x3 = 8 = 23
=> x = 2
\(\Leftrightarrow\frac{2+16}{9}=\frac{18}{9}=2\)
\(\Leftrightarrow\frac{y-25}{16}=2\) => y - 25 = 32 => y = 57
\(\Leftrightarrow\frac{z+9}{25}=2\) => z + 9 = 50 => z = 41
Vậy x = 2; y = 57; z = 41
\(2x^3-1=15\)
\(\Leftrightarrow2x^3=15+1=16\)
\(\Leftrightarrow x^3=\frac{16}{2}=8\)
\(\Leftrightarrow x=2\)
Thay \(x=2;\)ta có :
\(\frac{y-25}{16}=\frac{z+9}{25}=\frac{2+16}{9}=\frac{18}{9}\)
\(\Leftrightarrow\frac{y-25}{16}=\frac{z+9}{25}=2\)
\(\Rightarrow\hept{\begin{cases}\frac{y-25}{16}=2\\\frac{z+9}{25}=2\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}y-25=32\\z+9=50\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}y=57\\z=41\end{cases}}\)
Vậy ...
Đỗ Đức Đạt
Ta có:\(2x^3-1=15\Rightarrow x^3=8\Rightarrow x=2\)
\(\frac{y-25}{16}=2\Rightarrow y=2.16+25=57\)
\(\frac{z+9}{25}=2\Rightarrow z=25.2-9=41\)
\(2x^3-1=15\)
\(2x^3=16\)
\(x^3=8\)
\(x=2\)
\(\Rightarrow\frac{x+16}{9}=\frac{2+16}{9}=\frac{18}{9}=2\)
\(\Rightarrow\frac{y-25}{16}=2\)
\(\Rightarrow y-25=32\)
\(\Rightarrow y=57\)
\(\Leftrightarrow\frac{z+9}{25}=2\)
\(\Rightarrow z+9=50\)
\(\Rightarrow z=50-9=41\)
Vậy \(z=41;x=2;y=57\)