Tìm các số thực \(x\) và \(y\), biết:
a) \((3x - 2) + (2y + 1)i = (x + 1) - (y - 5)i\);
b) \((1 - 2x) - i\sqrt 3 = \sqrt 5 + (1 - 3y)i\);
c) \((2x + y) + (2y - x)i = (x - 2y + 3) + (y + 2x + 1)i\).
a) \((3x - 2) + (2y + 1)i = (x + 1) - (y - 5)i\) \(⇔\left\{\begin{matrix} 3x-2=x+1\\ 2y+1=-(y-5) \end{matrix}\right. ⇔ \left\{\begin{matrix} x=\frac{3}{2}\\ y=\frac{4}{3} \end{matrix}\right..\)
Vậy \( \left( {x;\;y} \right) = \left( {\frac{3}{2};\;\frac{4}{3}} \right).\)
b) \((1 - 2x) - i\sqrt 3 = \sqrt 5 + (1 - 3y)i\)
\( ⇔ \left\{\begin{matrix} 1-2x=\sqrt{5}\\ 1-3y=-\sqrt{3} \end{matrix}\right. ⇔ \left\{\begin{matrix} x=\frac{1-\sqrt{5}}{2}\\ y=\frac{1+\sqrt{3}}{3} \end{matrix}\right..\)
Vậy \( \left( {x;\;y} \right) = \left( \frac{1-\sqrt{5}}{2};\;\frac{1+\sqrt{3}}{3} \right).\)
c) \((2x + y) + (2y - x)i = (x - 2y + 3) + (y + 2x + 1)i\)
\( ⇔ \left\{\begin{matrix} 2x+y=x-2y+3\\ 2y-x=y+2x+1 \end{matrix}\right. ⇔ \left\{\begin{matrix} x+3y =3\\ -3x+y=1 \end{matrix}\right.\)
\(⇔ \left\{\begin{matrix} x=0\\ y=1 \end{matrix}\right.\).
Vậy \( \left( {x;\;y} \right)= \left( {0;\;1} \right).\)
a) \((3x - 2) + (2y + 1)i = (x + 1) - (y - 5)i\);
b) \((1 - 2x) - i\sqrt 3 = \sqrt 5 + (1 - 3y)i\);
c) \((2x + y) + (2y - x)i = (x - 2y + 3) + (y + 2x + 1)i\).
Lời giải bài tập
Từ định nghĩa bằng nhau của hai số phức, ta có:a) \((3x - 2) + (2y + 1)i = (x + 1) - (y - 5)i\) \(⇔\left\{\begin{matrix} 3x-2=x+1\\ 2y+1=-(y-5) \end{matrix}\right. ⇔ \left\{\begin{matrix} x=\frac{3}{2}\\ y=\frac{4}{3} \end{matrix}\right..\)
Vậy \( \left( {x;\;y} \right) = \left( {\frac{3}{2};\;\frac{4}{3}} \right).\)
b) \((1 - 2x) - i\sqrt 3 = \sqrt 5 + (1 - 3y)i\)
\( ⇔ \left\{\begin{matrix} 1-2x=\sqrt{5}\\ 1-3y=-\sqrt{3} \end{matrix}\right. ⇔ \left\{\begin{matrix} x=\frac{1-\sqrt{5}}{2}\\ y=\frac{1+\sqrt{3}}{3} \end{matrix}\right..\)
Vậy \( \left( {x;\;y} \right) = \left( \frac{1-\sqrt{5}}{2};\;\frac{1+\sqrt{3}}{3} \right).\)
c) \((2x + y) + (2y - x)i = (x - 2y + 3) + (y + 2x + 1)i\)
\( ⇔ \left\{\begin{matrix} 2x+y=x-2y+3\\ 2y-x=y+2x+1 \end{matrix}\right. ⇔ \left\{\begin{matrix} x+3y =3\\ -3x+y=1 \end{matrix}\right.\)
\(⇔ \left\{\begin{matrix} x=0\\ y=1 \end{matrix}\right.\).
Vậy \( \left( {x;\;y} \right)= \left( {0;\;1} \right).\)