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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
\(2x^3-1=15\)
\(\Leftrightarrow2x^3=15+1=16\)
\(\Leftrightarrow x^3=\frac{16}{2}=8\)
\(\Leftrightarrow x=2\)
Thay \(x=2;\)ta có :
\(\frac{y-25}{16}=\frac{z+9}{25}=\frac{2+16}{9}=\frac{18}{9}\)
\(\Leftrightarrow\frac{y-25}{16}=\frac{z+9}{25}=2\)
\(\Rightarrow\hept{\begin{cases}\frac{y-25}{16}=2\\\frac{z+9}{25}=2\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}y-25=32\\z+9=50\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}y=57\\z=41\end{cases}}\)
Vậy ...
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
Ta có:\(2x^3-1=15\Rightarrow x^3=8\Rightarrow x=2\)
\(\frac{y-25}{16}=2\Rightarrow y=2.16+25=57\)
\(\frac{z+9}{25}=2\Rightarrow z=25.2-9=41\)
\(2x^3-1=15\)
\(2x^3=16\)
\(x^3=8\)
\(x=2\)
\(\Rightarrow\frac{x+16}{9}=\frac{2+16}{9}=\frac{18}{9}=2\)
\(\Rightarrow\frac{y-25}{16}=2\)
\(\Rightarrow y-25=32\)
\(\Rightarrow y=57\)
\(\Leftrightarrow\frac{z+9}{25}=2\)
\(\Rightarrow z+9=50\)
\(\Rightarrow z=50-9=41\)
Vậy \(z=41;x=2;y=57\)