Từ \(\frac{9-x}{7}+\frac{11-x}{9}=2\)

\(=>\frac{9-x}{7}+\frac{11-x}{9}-2=0\)

\(=>\frac{9-x}{7}+\frac{11-x}{9}-1-1=0\)

\(=>\left(\frac{9-x}{7}-1\right)+\left(\frac{11-x}{9}-1\right)=0\)

\(=>\frac{2-x}{7}+\frac{2-x}{9}=0=>\left(2-x\right).\left(\frac{1}{7}+\frac{1}{9}\right)=0\)

Vì \(\frac{1}{7}+\frac{1}{9}\) khác 0=>2-x=0=>x=2

Theo T/c dãy tỉ số=nhau:

\(\frac{x+16}{9}=\frac{y-25}{16}=\frac{z+9}{25}=\frac{x+16+y-25+z+9}{9+16+25}\)\(=\frac{\left(x+y+z\right)+\left(16-25+9\right)}{9+16+25}=\frac{x+y+z}{50}\)

Thay x=2 vào \(\frac{x+16}{9}=>\frac{2+16}{9}=\frac{x+y+z}{50}=>\frac{x+y+z}{50}=2=>x+y+z=100\)

Vậy x+y+z=100

Ta có : \(\frac{9-x}{7}=\frac{11-x}{9}=1+\frac{2-x}{7}+1+\frac{2-x}{9}=2=>\left(2-x\right)\left(\frac{1}{7}+\frac{1}{9}\right)=0=>2-x=0=>x=2\)

Thế vào tìm đc y và z rồi ra x+y+z nha bạn 

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

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:  ...