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a)Ta có : 2x+2y-z-7=0 => 2x+2y-z=7
Ta có : \(x=\frac{y}{2}=>\frac{x}{2}=\frac{y}{4}\)
Mà \(\frac{y}{4}=\frac{z}{5}\)nên \(\frac{x}{2}=\frac{y}{4}=\frac{z}{5}=\frac{2x}{4}=\frac{2y}{8}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x}{2}=\frac{y}{4}=\frac{z}{5}=\frac{2x}{4}=\frac{2y}{8}=\frac{2x+2y-z}{4+8-5}=\frac{7}{7}=1\)
Từ \(\frac{x}{2}=1=>x=2\)
Từ\(\frac{y}{4}=1=>y=4\)
Từ \(\frac{z}{5}=1=>z=5\)
\(\frac{x}{2}=\frac{y}{4}=\frac{z}{5}=\frac{2x}{4}=\frac{2y}{8}\)

\(2x^3-1=15\)
\(\Leftrightarrow2x^3=15+1=16\)
\(\Leftrightarrow x^3=\frac{16}{2}=8\)
\(\Leftrightarrow x=2\)
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 ...

Ta có: \(\frac{x+16}{4}=\frac{4\left(x+16\right)}{4.4}=\frac{4x+64}{16}\)
Mà \(2x^3-1=15\)
\(\Rightarrow2x^3=15+1\)
\(\Rightarrow2x^3=16\)
\(\Rightarrow x^3=8\)
\(\Rightarrow x^3=2^3\)
\(\Rightarrow x=2\)
\(\Rightarrow\frac{x+16}{4}=\frac{2+16}{4}=\frac{18}{4}\)
Vì \(\frac{x+16}{4}=\frac{y-25}{16}\Rightarrow18.16=4\left(y-25\right)\)
\(\Rightarrow4y-100=288\)
\(\Rightarrow4y=388\)
\(\Rightarrow y=388:4\)
\(\Rightarrow y=97\)
\(\Rightarrow\frac{y-25}{16}=\frac{97-25}{16}=\frac{72}{16}\)
Tương tự: \(72.25=16\left(z+9\right)\)
\(\Rightarrow1800=16z+144\)
\(\Rightarrow16z=1800-144\)
\(\Rightarrow16z=1656\)
\(\Rightarrow z=1656:16\)
\(\Rightarrow z=103,5\)
Vậy: \(x+y+z=2+97+103,5=202,5\)

\(a,\frac{2x}{3}=\frac{2y}{4}=\frac{4z}{5}\)và x + y + z = 49
Ta có : \(\frac{2x}{3}=\frac{2y}{4}=\frac{4z}{5}=\frac{x}{\frac{3}{2}}=\frac{y}{\frac{4}{2}}=\frac{z}{\frac{5}{4}}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x}{\frac{3}{2}}=\frac{y}{\frac{4}{2}}=\frac{z}{\frac{5}{4}}=\frac{x+y+z}{\frac{3}{2}+\frac{4}{2}+\frac{5}{4}}=\frac{49}{\frac{19}{4}}=49\cdot\frac{4}{19}=\frac{196}{19}\)
Vậy : \(\hept{\begin{cases}\frac{x}{\frac{3}{2}}=\frac{196}{19}\\\frac{y}{\frac{4}{2}}=\frac{196}{19}\\\frac{z}{\frac{5}{4}}=\frac{169}{14}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{294}{19}\\y=\frac{392}{19}\\z=\frac{245}{19}\end{cases}}\)
\(b,\frac{x}{y}=\frac{3}{4};\frac{y}{z}=\frac{5}{7}\)và 2x + 3y - z = 186
Ta có : \(\frac{x}{y}=\frac{3}{4};\frac{y}{z}=\frac{5}{7}\Leftrightarrow\frac{x}{3}=\frac{y}{4};\frac{y}{5}=\frac{z}{7}\)
\(\Leftrightarrow\frac{x}{15}=\frac{y}{20};\frac{y}{20}=\frac{z}{28}\)
\(\Leftrightarrow\frac{x}{15}=\frac{y}{20}=\frac{z}{28}\)
\(\Leftrightarrow\frac{2x}{30}=\frac{3y}{60}=\frac{z}{28}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{2x}{30}=\frac{3y}{60}=\frac{z}{28}=\frac{2x+3y-z}{30+60-28}=\frac{186}{62}=3\)
Vậy : \(\hept{\begin{cases}\frac{x}{15}=3\\\frac{y}{20}=3\\\frac{z}{28}=3\end{cases}}\Leftrightarrow\hept{\begin{cases}x=45\\y=60\\z=84\end{cases}}\)

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

Theo đề bài, ta có:
\(2x^3-1=15\)
\(\Rightarrow2x^3=15+1\)
\(\Rightarrow2x^3=16\)
\(\Rightarrow x^3=16\div2\)
\(\Rightarrow x^3=8\)
\(\Rightarrow x^3=2^3\)
\(\Rightarrow x=2\)
\(\Rightarrow\frac{x+16}{9}=\frac{2+16}{9}=\frac{18}{9}=2\) (1)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{x+16}{9}=\frac{y-25}{16}=\frac{z+9}{25}=\frac{x+16+y-25+z+9}{9+16+25}=\frac{\left(x+y+z\right)+\left(16-25+9\right)}{9+16+25}=\frac{\left(x+y+z\right)+0}{50}=\frac{x+y+z}{50}\) (2)
Từ (1) và (2) \(\Rightarrow\frac{x+y+z}{50}=2\)
\(\Rightarrow x+y+z=2\times50\)
\(\Rightarrow x+y+z=100\)
Vậy giá trị tổng \(x+y+z\) bằng 100.

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
\(\left|\frac{1}{15}-x\right|+\left|\frac{2}{25}-y\right|+\left|z-\frac{16}{5}\right|=0\)
Ta có: \(\hept{\begin{cases}\left|\frac{1}{15}-x\right|\ge0\forall x\\\left|\frac{2}{25}-y\right|\ge0\forall y\\\left|z-\frac{16}{5}\right|\ge0\forall z\end{cases}}\)\(\Rightarrow\left|\frac{1}{15}-x\right|+\left|\frac{2}{25}-y\right|+\left|z-\frac{16}{5}\right|\ge0\forall x;y;z\)
Mà \(\left|\frac{1}{15}-x\right|+\left|\frac{2}{25}-y\right|+\left|z-\frac{16}{5}\right|=0\)
\(\Rightarrow\hept{\begin{cases}\left|\frac{1}{15}-x\right|=0\\\left|\frac{2}{25}-y\right|=0\\\left|z-\frac{16}{5}\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{15}\\y=\frac{2}{25}\\z=\frac{16}{5}\end{cases}}\)
Vậy \(\hept{\begin{cases}x=\frac{1}{15}\\y=\frac{2}{25}\\z=\frac{16}{5}\end{cases}}\)