Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
10,\\
\dfrac{{a + b}}{{a - b}} = \dfrac{{c + a}}{{c - a}}\\
\Leftrightarrow \left( {a + b} \right)\left( {c - a} \right) = \left( {c + a} \right)\left( {a - b} \right)\\
\Leftrightarrow ac - {a^2} + bc - ab = ca - bc + {a^2} - ab\\
\Leftrightarrow - {a^2} + bc = - bc + {a^2}\\
\Leftrightarrow 2bc = 2{a^2}\\
\Leftrightarrow {a^2} = bc\\
11,\\
x:y:z = 3:5:\left( { - 2} \right)\\
\Leftrightarrow \dfrac{x}{3} = \dfrac{y}{5} = \dfrac{z}{{ - 2}}\\
\Leftrightarrow \dfrac{{5x}}{{5.3}} = \dfrac{y}{5} = \dfrac{{3z}}{{3.\left( { - 2} \right)}}\\
\Leftrightarrow \dfrac{{5x}}{{15}} = \dfrac{y}{5} = \dfrac{{3z}}{{ - 6}} = \dfrac{{5x - y + 3z}}{{15 - 5 + \left( { - 6} \right)}} = \dfrac{{124}}{4} = 31\\
\Rightarrow \left\{ \begin{array}{l}
\dfrac{{5x}}{{15}} = 31\\
\dfrac{y}{5} = 31\\
\dfrac{{3z}}{{ - 6}} = 31
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\dfrac{x}{3} = 31\\
\dfrac{y}{5} = 31\\
\dfrac{z}{{ - 2}} = 31
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 93\\
y = 155\\
z = - 62
\end{array} \right.
\end{array}\)
