Quá đơn giản :

2x³-1 = 15

=> 2x³=16

=> x³ = 8

=> x  =2

Thay x vào \(\frac{x+16}{9}\)

=> \(\frac{2+16}{9}=2\)

=> \(2=\frac{y-25}{16}\)

=> y-25 = 32

=> y = 57

=> \(2=\frac{z+9}{25}\)

=> z + 9 = 50

=> ...

Đ/S:  ...

Đặt \(\frac{x-16}{9}=\frac{y-25}{16}=\frac{x+9}{25}=k\)

\(\Rightarrow\hept{\begin{cases}x=9k+16\\y=16k+25\\z=25k-9\end{cases}}\Rightarrow x+y+z=9k+16+16k+25+25k-9=50k+32\)

Xét 2x³-1=15

=>2x³=15+1=16

=>x³=16:2=8

=>x³=2³

=>x=2

=>k=\(\frac{2-16}{9}=\frac{-14}{9}\)

=>x+y+z=\(50.\frac{-14}{9}+32=\frac{-412}{9}\)

Do 2x^3-1=15= =>2x^3=16 =>x^3m=8 =>x=2

ta co x-16/9 =y-25/16 =z+9/25 =x-16+y-25+z+25/   9+16+25  =x+y+z-32 /50  (1)

thay x=2 vao (1) ta co 2-16/9= -14/9=x+y+z-32 /50 =>x+y+z-32 =-700/9

=>x+y+z=-412/9

2x^3-1=15 => 2x^3 = 15+1 = 16

=> x^3=16:2=8 = 2^3

=> x=2

Khi đó : y-25/16=z+9/25=x+16/9 = 2+16/9 = 2

=> y = 57 ; z = 41

=> x+y+z = 2+57+41 = 100

k mk nha

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

=> 2x^3 = 15+1 = 16

=> x^3 = 16:2 = 8 = 2^3

=> x = 2

=> y-25/16 = z+9/25 = 2+16/9 = 2

=> y = 57 ; z = 41

=> x+y+z = 2+57+41 = 100

Vậy x+y+z = 100

Tk mk nha

Ta có : 2x³ - 1 = 15 \(\Rightarrow\)2x³ = 16 \(\Rightarrow\)x³ = 8 = 2³ \(\Rightarrow\)x = 2

Thay x = 2 vào các tỉ số trên, ta được :

\(\frac{2+16}{9}=\frac{y-25}{17}=\frac{z+9}{25}\)

hay \(\frac{y-25}{17}=\frac{z+9}{25}=2\)

\(\frac{y-25}{17}=2\Rightarrow y-25=34\Rightarrow y=59\)

\(\frac{z+9}{25}=2\Rightarrow z+9=50\Rightarrow z=41\)

Vậy x + y + z = 2 + 59 + 41 = 102

\(a,\frac{2x}{3}=\frac{2y}{4}=\frac{4z}{5}\)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 : \(\hept{\begin{cases}\frac{x}{15}=3\\\frac{y}{20}=3\\\frac{z}{28}=3\end{cases}}\Leftrightarrow\hept{\begin{cases}x=45\\y=60\\z=84\end{cases}}\)

\(c,x:y:z=12:9:5\)và xyz = 20

Ta có : \(x:y:z=12:9:5\)hay \(\frac{x}{12}=\frac{y}{9}=\frac{z}{5}\)

Đặt : \(\frac{x}{12}=\frac{y}{9}=\frac{z}{5}=k\)

\(\Leftrightarrow\hept{\begin{cases}x=12k\\y=9k\\z=5k\end{cases}}\)

\(\Leftrightarrow xyz=12k\cdot9k\cdot5k=540k^3\)

\(\Leftrightarrow xyz=540k^3\)

\(\Leftrightarrow540k^3=xyz\)

\(\Leftrightarrow540k^3=20\)

\(\Leftrightarrow k^3=\frac{20}{540}=\frac{1}{27}\Leftrightarrow k=\frac{1}{3}\)

Vậy : \(\hept{\begin{cases}x=12\cdot\frac{1}{3}\\y=9\cdot\frac{1}{3}\\z=5\cdot\frac{1}{3}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=4\\y=3\\z=\frac{5}{3}\end{cases}}\)

2+57+41=100