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Ta có: \(\frac{x+16}{4}=\frac{4\left(x+16\right)}{4.4}=\frac{4x+64}{16}\)
Mà \(2x^3-1=15\)
\(\Rightarrow2x^3=15+1\)
\(\Rightarrow2x^3=16\)
\(\Rightarrow x^3=8\)
\(\Rightarrow x^3=2^3\)
\(\Rightarrow x=2\)
\(\Rightarrow\frac{x+16}{4}=\frac{2+16}{4}=\frac{18}{4}\)
Vì \(\frac{x+16}{4}=\frac{y-25}{16}\Rightarrow18.16=4\left(y-25\right)\)
\(\Rightarrow4y-100=288\)
\(\Rightarrow4y=388\)
\(\Rightarrow y=388:4\)
\(\Rightarrow y=97\)
\(\Rightarrow\frac{y-25}{16}=\frac{97-25}{16}=\frac{72}{16}\)
Tương tự: \(72.25=16\left(z+9\right)\)
\(\Rightarrow1800=16z+144\)
\(\Rightarrow16z=1800-144\)
\(\Rightarrow16z=1656\)
\(\Rightarrow z=1656:16\)
\(\Rightarrow z=103,5\)
Vậy: \(x+y+z=2+97+103,5=202,5\)
\(2x^3-1=15\)
\(\Leftrightarrow2x^3=15+1=16\)
\(\Leftrightarrow x^3=\frac{16}{2}=8\)
\(\Leftrightarrow x=2\)
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 ...
\(\frac{x+16}{9}=\frac{y-25}{16}=\frac{z+9}{25}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x+16}{9}=\frac{y-25}{16}=\frac{z+9}{25}=\frac{2\left(x+16\right)+3\left(y-25\right)-\left(z+9\right)}{2.9+3.16-25}\)
\(=\frac{2x+3y-z-52}{41}=\frac{50-52}{41}=\frac{-2}{41}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{-2}{41}.9-16=\frac{-674}{41}\\y=\frac{-2}{41}.16+25=\frac{993}{41}\\z=-\frac{2}{41}.25-9=\frac{-419}{41}\end{cases}}\)
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
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
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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