Question

X'2+y'2=50 xy=7 решить систему

Answer 1

{x^2 + y^2 = 50
{xy = 7
y=7/x
x^2 + 49/x^2 = 50
(x^4 - 50x^2 + 49)/x^2=0
Пусть х^2=t
t^2 - 50t + 49 = 0
t1=1 ; t2=49
x^2=1 ; x^2=49
x=1 ; x=7
y=7/x
{x=1 {x=7
{y=7 {y=1
Ответ: (1;7) ; (7;1).

Answer 2

Ответ:

(1;7), (7;1) , (-1;-7) , (-7;-1).

Объяснение:

$left { begin{array}{lcl} {{x^{2}+y^{2}=50,  } \ {xy=7;}} end{array} right. Leftrightarrowleft { begin{array}{lcl} {{x^{2}+y^{2} =50, } \ {2xy=14;}} end{array} right.Leftrightarrowleft { begin{array}{lcl} {{x^{2}+2xy+y^{2} =64, } \ {xy=7;}} end{array} right. Leftrightarrow$

$left { begin{array}{lcl} {{(x+y)^{2} =64,} \ {xy=7}} end{array} right.Leftrightarrowleft { begin{array}{lcl} {{left [ begin{array}{lcl} {{x+y=8,} \ {x+y=-8,}} end{array} right.} \ {xy=7.}} end{array} right.$

Решим первую систему:

$left { begin{array}{lcl} {{x+y=8,} \ {xy=7;}} end{array} right.Leftrightarrowleft { begin{array}{lcl} {{y=8-x,} \ {x(8-x) =7;}} end{array} right.Leftrightarrow$

$left { begin{array}{lcl} {{y=8-x,} \ {x^{2} -8x+7=0;}} end{array} right.Leftrightarrowleft { begin{array}{lcl} {{y=8-x} \ {left [begin{array}{lcl} {{x=1} \ {x=7}} end{array} right.}} end{array}$$Leftrightarrowleft [ begin{array}{lcl} {left { begin{array}{lcl} {{x=1,} \ {y=7;}} end{array} right{} \ {left { begin{array}{lcl} {{x=7,} \ {y=1;}} end{array} right}} end{array} right$

Решим вторую систему:

$left { begin{array}{lcl} {{x+y=-8,} \ {xy=7;}} end{array} right.Leftrightarrowleft { begin{array}{lcl} {{y=-8-x} \ {x(-8-x)=7;}} end{array} right.Leftrightarrow left { begin{array}{lcl} {{y=-8-x,} \ {x^{2} +x+7=0;}} end{array} right. Leftrightarrow$

$left { begin{array}{lcl} {{y=-8-x,} \ {left [ begin{array}{lcl} {{x=-1,} \ {x=-7;}} end{array} right.}} end{array} right.  Leftrightarrow  left [ begin{array}{lcl} {{left { begin{array}{lcl} {{x=-1,} \ {y=-7;}} end{array} right.} \ {left { begin{array}{lcl} {{x=-7,} \ {y=-1.}} end{array} right.}} end{array} right.$