Bài 1:Tìm x,y x/3=y/5 và x.y = 60 Bài 2 2x/3 = 3y/4 = 4z/5 và x+y+z=49

Đáp án:

B2:

\(\left\{ \begin{array}{l}
y = 16\\
x = 18\\
z = 15
\end{array} \right.\)

Giải thích các bước giải:

\(\begin{array}{l}
B1:\\
\left\{ \begin{array}{l}
x = \dfrac{3}{5}y\\
\dfrac{3}{5}y.y = 60
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
x = \dfrac{3}{5}y\\
{y^2} = 100
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
x = \dfrac{3}{5}y\\
\left| y \right| = 10
\end{array} \right.\\
 \to \left[ \begin{array}{l}
y = 10\\
y =  – 10
\end{array} \right. \to \left[ \begin{array}{l}
x = 6\\
x =  – 6
\end{array} \right.\\
B2:\\
\left\{ \begin{array}{l}
x = \dfrac{9}{8}y\\
z = \dfrac{{15}}{{16}}y\\
x + y + z = 49
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
x = \dfrac{9}{8}y\\
z = \dfrac{{15}}{{16}}y\\
\dfrac{9}{8}y + y + \dfrac{{15}}{{16}}y = 49
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
x = \dfrac{9}{8}y\\
z = \dfrac{{15}}{{16}}y\\
\left( {\dfrac{9}{8} + 1 + \dfrac{{15}}{{16}}} \right)y = 49
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
x = \dfrac{9}{8}y\\
z = \dfrac{{15}}{{16}}y\\
\dfrac{{49}}{{16}}y = 49
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
y = 16\\
x = 18\\
z = 15
\end{array} \right.
\end{array}\)