Đáp án:
a, \(\left( {x;y;z} \right) = \left( {2; - 7;1} \right)\)
b, \(\left( {x;y;z} \right) = \left( {6;4;3} \right)\)
Giải thích các bước giải:
a,
Ta có:
\(\begin{array}{l}
4{x^3} - 3 = 29\\
\Leftrightarrow 4{x^3} = 32\\
\Leftrightarrow {x^3} = 8\\
\Leftrightarrow x = 2\\
\frac{{x + 16}}{9} = \frac{{y - 25}}{{ - 16}} = \frac{{z + 49}}{{25}}\\
\Leftrightarrow \frac{{2 + 16}}{9} = \frac{{y - 25}}{{ - 16}} = \frac{{z + 49}}{{25}}\\
\Leftrightarrow \frac{{y - 25}}{{ - 16}} = \frac{{z + 49}}{{25}} = 2\\
\Leftrightarrow \left\{ \begin{array}{l}
\frac{{y - 25}}{{ - 16}} = 2\\
\frac{{z + 49}}{{25}} = 2
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
y - 25 = - 32\\
z + 49 = 50
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
y = - 7\\
z = 1
\end{array} \right. \Rightarrow \left( {x;y;z} \right) = \left( {2; - 7;1} \right)
\end{array}\)
b,
\(\begin{array}{l}
2x = 3y = 4z\\
\Leftrightarrow \frac{{2x}}{{12}} = \frac{{3y}}{{12}} = \frac{{4z}}{{12}}\left( {12 = BCNN\left( {2;3;4} \right)} \right)\\
\Rightarrow \frac{x}{6} = \frac{y}{4} = \frac{z}{3}
\end{array}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\begin{array}{l}
\frac{x}{6} = \frac{y}{4} = \frac{z}{3} = \frac{{x + y + z}}{{6 + 4 + 3}} = \frac{{13}}{{13}} = 1\\
\Rightarrow \left\{ \begin{array}{l}
\frac{x}{6} = 1\\
\frac{y}{4} = 1\\
\frac{z}{3} = 1
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x = 6\\
y = 4\\
z = 3
\end{array} \right. \Rightarrow \left( {x;y;z} \right) = \left( {6;4;3} \right)
\end{array}\)
