\(\frac{9}{25}\) x 3,14=
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Ta có \(\frac{9}{25}x+\frac{3}{5}.\frac{9}{25}x+\frac{3}{5}.18+\frac{9}{25}x+18=x\)
=> \(\frac{9}{25}x\left(1+\frac{3}{5}+1\right)+18\left(\frac{3}{5}+1\right)=x\)
=> \(\frac{117}{125}x+28,8=x\)
=> \(x-\frac{117}{125}x=28,8\)
=> \(\frac{8}{125}x=28,8\)
=> x = 450
Vậy x = 450
\(\frac{9}{25}x+\frac{3}{5}.\frac{9}{25}x+\frac{3}{5}.18+\frac{9}{25}x+18=x\)
\(x\left(\frac{9}{25}+\frac{9}{25}+\frac{9}{25}\right).\frac{3}{5}+\frac{3}{5}.18+18=x\)
\(x.\frac{3}{5}\left(\frac{27}{25}+18\right)+18=x\)
\(x.\frac{3}{5}\left(\frac{27}{25}+\frac{450}{25}\right)+18=x\)
\(x.\frac{3}{5}.\frac{477}{25}+18=x\)
\(x.\frac{1431}{125}+\frac{2250}{125}=x\)
\(x.\frac{3681}{125}=x\)
vậy chac tui làm sai rồi
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