Bài 4.1 trang 198 SBT giải tích 12

Đề bài

Tìm các số thực \(x, y\) thỏa màn:

a) \(2x + 1 + (1 – 2y)i\) \( = 2 – x + (3y – 2)i\)

b) \(4x + 3 + (3y – 2)i \) \( = y  +1 + (x – 3)i\)

c) \(x + 2y + (2x – y)i \) \( = 2x + y + (x  + 2y)i\)

Lời giải chi tiết

a) Ta có: \(\left( {2x + 1} \right) + \left( {1 - 2y} \right)i\) \( = \left( {2 - x} \right) + \left( {3y - 2} \right)i\)

\( \Leftrightarrow \left\{ \begin{array}{l}2x + 1 = 2 - x\\1 - 2y = 3y - 2\end{array} \right.\) \( \Leftrightarrow \left\{ \begin{array}{l}3x = 1\\5y = 3\end{array} \right.\) \( \Leftrightarrow \left\{ \begin{array}{l}x = \dfrac{1}{3}\\y = \dfrac{3}{5}\end{array} \right.\)

Vậy \(x = \dfrac{1}{3},y = \dfrac{3}{5}\)

b) \(4x + 3 + \left( {3y - 2} \right)i\) \( = {\rm{ }}y\; + 1 + \left( {x - 3} \right)i\)

\( \Leftrightarrow \left\{ \begin{array}{l}4x + 3 = y + 1\\3y - 2 = x - 3\end{array} \right.\) \( \Leftrightarrow \left\{ \begin{array}{l}4x - y =  - 2\\x - 3y = 1\end{array} \right.\) \( \Leftrightarrow \left\{ {} \right.\)

Vậy \(x =  - \dfrac{7}{{11}},y =  - \dfrac{6}{{11}}\).

c) \(x + 2y + \left( {2x - y} \right)i\) \( = 2x + y + \left( {x + 2y} \right)\)

\( \Leftrightarrow \left\{ \begin{array}{l}x + 2y = 2x + y\\2x - y = x + 2y\end{array} \right.\) \( \Leftrightarrow \left\{ \begin{array}{l}x - y = 0\\x - 3y = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = 0\\y = 0\end{array} \right.\)

Vậy \(x = y = 0\).