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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
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\(2x^3-1=15\)
\(\Leftrightarrow2x^3=15+1=16\)
\(\Leftrightarrow x^3=\frac{16}{2}=8\)
\(\Leftrightarrow x=2\)
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 ...
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Ta có 2x3 - 1 = 15
2x3 = 15+ 1=16
x3 =16:2=8 =>x3=23 => x=2
Thay x = 2 vào biểu thức\(\frac{x+16}{9}\)=> \(\frac{2+16}{9}=\frac{18}{9}=2\)
*\(\frac{y-25}{16}=2\)=> y-25=32 => y = 57
*\(\frac{z+9}{25}=2\)=>z+ 9 = 50 => z= 50-9 = 41
vậy x + y +z = 2 + 57 + 41 = 100
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Theo giả thiết thứ nhất, ta có:
x+16/9=y−25/16=z+9/25=x+16+y−25+z+9/9+16+25=x+y+z/50
Theo giả thiết thứ 2:
9−x/7+11−x/9=2.
=>81−9x/63+77−7x/63=81−9x+77−7x/63=158−16x/63=2
=>158−16x=2.63=126=>x=2
Theo giả thiết thứ nhất:
x+16/9=x+y+z/50
=>2+16/9=x+y+z/50
=>x+y+z=2.50=100
Vậy x+y+z=100