Question

решите систему уравнений​

Answer 1

Объяснение:

{2х-у= -1 2х+1=у

{5х-у²= -4

5х-(2х+1)²= -4

5х-(4х²+4х+1)= -4

5х-4х²-4х-1= -4

-4х²+х-1+4=0

-4х²+х+3=0

4х²-х-3=0

D=(-1)²-4×4×(-3)=1+48=49

x1=(1+7)/2×4=1

x2=(1-7)/8= -6/8= -3/4

y1=2×1+1=3

y2=2×(-3/4)+1= -3/2+2/2= -1/2

ответ: (1;3) (-3/4;-1/2)

Answer 2

***

$\displaystyle \bf \left \{ {{2x-y=-1 \ \ | \cdot (-1)} \atop {5x-y^2=-4 \ | \cdot (-1)}} \right.$

$\displaystyle \bf \left \{{{y-2x=1} \atop {y^2-5x=4}} \right. \ \ \ \ \ \left \{ {{y=1+2x} \atop {(1+2x)^2-5x=4}} \right. \ \ \ \ \ \left \{ {{y=1+2x} \atop {4x^2+4x+1-5x=4}} \right.$

$\displaystyle \bf \left \{ {{y=1+2x} \atop {4x^2+4x-5x=4-1}} \right.$

$\displaystyle \bf 4x^2 +4x-5x=4-1\\\\4x^2+4x-5x=3\\\\4x^2-x-3=0\\\\\\D=b^2-4ac = \Big(-1\Big)^2-4 \cdot 4 \cdot \Big(-4\Big)=1-4\cdot \:4\cdot \Big(-3\Big)=1-\left(-48\right)=1+48=49\\\\\\X_{1:2}=\frac{-b\pm \sqrt D}{2a} = \frac{1\pm\sqrt {49}}{2\cdot 4} =\frac{1\pm7}{8}$

$\displaystyle \bf X_1=\frac{1+7}{8} =\frac{8}{8} =1\\\\X_2=\frac{1-7}{8} =\frac{-6}{8} =-0,75$

$\displaystyle \bf \left \{ {{y=1+2x} \atop {\begin{bmatrix}x=1\\ x=-0,75\end{m}} \right. \\\\\ \\\\\ \ \ \left \{ {{y=1+2x} \atop {x=1}} \right. \ \ \ \cup \ \ \ \left \{ {{y=1+2x} \atop {x=-0,75}} \\\right. \\\\\\\left \{ {{y=1+2\cdot 1} \atop {x=1}} \right. \ \ \ \cup \ \ \ \left \{ {{y=1-1.5} \atop {x=-0,75}} \right. \\\\\\\left \{ {{y=3} \atop {x=1}} \right. \ \ \ \ \ \ \ \ \ \ \ \cup \ \ \ \ \ \ \ \ \ \ \left \{ {{y=-0,5} \atop {x=-,75}} \right.$

ответ:

$\boxed{\displaystyle \bf \Big \ \Big(-0,75; \ -0,5\Big) \ \ \cup \ \ \Big(1;3\Big)}$