Question

Cho \(\frac{x+16}{9}=\frac{y-25}{16}=\frac{z+9}{25}\) và \(\frac{9-x}{7}+\frac{11-x}{9}=2\).Tính x+y+z?

Answer 1

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