Có: 

Có: 

Ta có:

Ta có: \(2x^3-1=15\Leftrightarrow x^3=8\Rightarrow x=2\)

\(\Rightarrow\dfrac{18}{9}=\dfrac{y-25}{16}=\dfrac{z+9}{25}\Rightarrow\left\{{}\begin{matrix}\dfrac{y-25}{16}=2\Rightarrow y=57\\\dfrac{z+9}{25}=2\Rightarrow z=41\end{matrix}\right.\)

Vậy \(B=x+y+z=2+57+41=100\)

`2x^3-1=15=>2x^3=16=>x^3=8=>x=2`

Có:`[x+16]/9=[y-25]/16`

`=>[2+16]/9=[y-25]/16=>y=57`

Có:`[x+16]/9=[z+9]/25`

`=>[2+16]/9=[z+9]/25=>z=41`

Ta có:`B=x+y+z=2+57+41=100`

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

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:  ...

Vậy \(x+y+z=100.\)

Chúc bạn học tốt!

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