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 $x = y = 0$.

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