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 

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:  ...

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

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