Tìm \(x, y, z\) biết:
a) \(x + 1 = - 2\)
b) \(x:2 = 10:5\)
c) \({\rm{x:2 = y:3}}\) và\({\rm{x + y = 10}}\)
d) \(3x = 2y; 7y = 5z\) và \(x – y + z = 32\)
Giải chi tiết:
\(\begin{array}{l}a) x + 1 = - 2\\x = - 2 - 1\\x = - 3\end{array}\)
\(\begin{array}{l}b) x:2 = 10:5\\x:2 = 2\\x = 2.2\\x = 4\end{array}\)
\(\begin{array}{l} c)x:2 = y:3 \Rightarrow \frac{x}{2} = \frac{y}{3} = \frac{{x + y}}{{2 + 3}} = \frac{{10}}{5} = 2\\+ )\frac{x}{2} = 2 \Rightarrow x = 2.2 = 4\\+ )\frac{y}{3} = 2 \Rightarrow y = 3.2 = 6\end{array}\)
Vậy \(x = 4;y = 6\).
d) Ta có:
\(3x = 2y;7y = 5z \Rightarrow \frac{x}{2} = \frac{y}{3};\frac{y}{5} = \frac{z}{7} \Rightarrow \frac{x}{{10}} = \frac{y}{{15}} = \frac{z}{{21}}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\begin{array}{l}\frac{x}{{10}} = \frac{y}{{15}} = \frac{z}{{21}} = \frac{{x - y + z}}{{10 - 15 + 21}} = \frac{{32}}{{16}} = 2\\+ )\frac{x}{{10}} = 2 \Rightarrow x = 10.2 = 20\\+ )\frac{y}{{15}} = 2 \Rightarrow x = 15.2 = 30\\+ )\frac{z}{{21}} = 2 \Rightarrow x = 21.2 = 42\end{array}\)
Vậy \(x = 20;y = 30;z = 42\).
chọn A.