Verified answer

Ответ:

1)

[tex]\displaystyle \left \{ {{ x = 0,4 } \atop { y = -\dfrac{3}{7} }} \right.[/tex]

2)

[tex]\left [ \begin{array}{l} \displaystyle \left \{ {{ x = -\dfrac{1}{6} } \atop { y = 2 }} \right \\ \\\displaystyle \left \{ {{ x = -\dfrac{1}{6} } \atop { y =-2 }} \right. \end{array} \right[/tex]

3)

[tex]\left [ \begin{array}{l} \displaystyle \left \{ {{ x = 3 } \atop { y = 1}} \right \\ \\\displaystyle \left \{ {{ x = -\dfrac{7}{3} } \atop { y = -\dfrac{5}{3} }} \right. \end{array} \right[/tex]

Примечание:

Комплексным числом называют число вида:

[tex]z = a +bi[/tex], где [tex]a,b \in \mathbb R[/tex], а число [tex]i[/tex] - называется мнимой единицей и для которой верное тождество [tex]i^{2} = -1[/tex]

Два комплексных числа [tex]z_{1} = a_{1} + b_{1}i[/tex] и [tex]z_{2} = a_{2} + b_{2}i[/tex] называют равными [tex](z_{1} = z_{2})[/tex] если:

[tex]\displaystyle[/tex][tex]\displaystyle \left \{ {{a_{1} = a_{2}} \atop {b_{1}=b_{2}}} \right.[/tex]

Пошаговое объяснение:

1)

[tex]5x - 7yi = 2 + 3i[/tex]

[tex]\displaystyle \left \{ {{5x =2 |:5} \atop {-7y = 3|:(-7)}} \right \left \{ {{x= 0,4} \atop { y = -\dfrac{3}{7} }} \right.[/tex]

2)

[tex]6x - y^{2}i = -1 -4i[/tex]

[tex]\displaystyle \left \{ {{6x = -1|:6} \atop { -y^{2} = -4 | \cdot (-1) }} \right \left \{ {{x = -\dfrac{1}{6} } \atop { y^{2} = 4 }} \right[/tex]

[tex]y^{2} = 4[/tex]

[tex]\sqrt{y^{2}} = \sqrt{4}[/tex]

[tex]|y| = 2[/tex]

[tex]y_{1,2} = \pm 2[/tex]

[tex]x = -\dfrac{1}{6}[/tex]

3) [tex](x + 3y^{2}i) - (2y - xi) = 1 + 6i[/tex]

[tex]x + 3y^{2}i - 2y + xi = 1 + 6i[/tex]

[tex]x - 2y+ 3y^{2}i + xi = 1 + 6i[/tex]

[tex]x - 2y+ i(3y^{2} + x) = 1 + 6i[/tex]

[tex]\displaystyle \left \{ {{x - 2y = 1} \atop { 3y^{2} + x = 6}} \right \left \{ {{x = 1 + 2y} \atop { 3y^{2} + x = 6}} \right \Longrightarrow 3y^{2} + 1 + 2y = 6[/tex]

[tex]3y^{2} + 1 + 2y = 6[/tex]

[tex]3y^{2} + 2y - 5= 0[/tex]

[tex]D = 4 - 4 \cdot 3 \cdot (-5) = 4 + 60 = 64 = 8^{2}[/tex]

[tex]y_{1} = \dfrac{-2 + 8}{2 \cdot 3} = \dfrac{6}{6} = 1[/tex]

[tex]y_{2} = \dfrac{-2 - 8}{2 \cdot 3} = \dfrac{-10}{6} = -\dfrac{5}{3}[/tex]

[tex]x_{1} = 1 + 2y_{1} = 1 + 2 \cdot 1 = 1 + 2 = 3[/tex]

[tex]x_{2} = 1 + 2y_{2} = 1 + 2 \cdot \bigg( -\dfrac{5}{3} \bigg) = 1 - \dfrac{10}{3} =\dfrac{3 - 10}{3} = -\dfrac{7}{3}[/tex]