Tìm x :
2x - 25 = 15 + 16
Tìm x,y,z biết:
x+16/9 = y-25/16 = z+9/25 và 2x2-1=15
Cho : (x+6)/9 = (y-25)/16 = (z+9)/25. Biết 2x^3 -1 = 15
Tìm x, y,z
Cho 3x-2y/4 = 2z-4x/3 = 4y-3z/2. CMR: x/2 = y/3 = z/4
Cho x+16/9 = y-25/16 = z+9/25 và 2x^3-1 = 15. Tìm x, y, z
Cho \(\frac{x+16}{9}=\frac{y-25}{16}=\frac{z+9}{25}\) và \(2x^3-1=15\)
Tìm x,y,z.
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
Cho \(\frac{x+16}{9}=\frac{y-25}{16}=\frac{z+9}{25}\) và \(2x^3-1=15.\)Tìm \(x+y+z\)
\(2x^3-1=15\)
\(\Leftrightarrow2x^3=15+1=16\)
\(\Leftrightarrow x^3=\frac{16}{2}=8\)
\(\Leftrightarrow x=2\)
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 ...
x+16/9=y-25/16=z+9/25 và 2x^3-1=15
Cho: \(\frac{x+16}{9}=\frac{y-25}{16}=\frac{z+9}{25}\) và\(2x^3-1=15\).Tìm x+y+z.
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 \(z=41;x=2;y=57\)
Cho (x+16)/9=(y-25)/16=(z+9)/25 và (2x^3)-1=15 Tính x+y+z
2x3−1=15⇒2x3=16⇒x3=8⇒x=22x3-1=15⇒2x3=16⇒x3=8⇒x=2
Có: x+169=y−2516x+169=y-2516
⇒2+169=y−2516⇒y=57⇒2+169=y-2516⇒y=57
Có: x+169=z+925x+169=z+925
⇒2+169=z+925⇒z=41⇒2+169=z+925⇒z=41
Ta có:B=x+y+z=2+57+41=100
Cho x+16/9=y-25/16=z+9/25 và 2x^3-1=15 . tính x+y+z = ?