Question

cho \(\dfrac{x+16}{9}=\dfrac{y-25}{16}=\dfrac{z+9}{25}\) và \(2x^3-1=15\) Tính B=x+y+z

Answer 1

Ta có :

\(2x^3-1=15\)

\(\Leftrightarrow2x^3=16\)

\(\Leftrightarrow x^3=8\)

\(\Leftrightarrow x=2\)

Thay \(x=2\) zô : \(\dfrac{x+16}{9}=\dfrac{y-25}{16}=\dfrac{z+9}{25}\)

\(\Leftrightarrow\dfrac{2+16}{9}=\dfrac{y-25}{16}=\dfrac{z+9}{25}\)

\(\Leftrightarrow\dfrac{y-25}{16}=\dfrac{z+9}{25}=\dfrac{18}{9}=2\)

+) \(\dfrac{y-25}{16}=2\)

\(\Leftrightarrow y-25=32\)

\(\Leftrightarrow y=57\)

+) \(\dfrac{z+9}{25}=2\)

\(\Leftrightarrow z+9=50\)

\(\Leftrightarrow z=41\)

Ta có :

\(\left\{{}\begin{matrix}x=2\\y=57\\z=41\end{matrix}\right.\) \(\Leftrightarrow x+y+z=2+57+41=100\)