Question

rezolvați sistemele x²-3x=y x-y=1

Answer 1

x²-3x=y

x-y=1 => y=x-1

x²-3x=x-1

x²-4x+1=0

D=16-4=12

x1=(4+2√3)/2

x1=2+√3 =>y1=1+√3

x2=(4-2√3)/2

x2=2-√3 =>y2=1-√3

(x,y)∈{(2+√3, 1+√3), (2-√3, 1-√3)}

Answer 2

$\left \{ {{ x^{2} -3x=y} \atop {x-y=1}} \right. \\ \\ \left \{ {{ x^{2} -3x=y} \atop {x-1=y}} \right. \\ \\ \left \{ {{ x^{2} -3x=x-1} \atop {y=x-1}} \right. \\ \\ \left \{ {{ x^{2} -3x-x+1=0} \atop {x-y=1}} \right. \\ \\ \left \{ {{ x^{2} -4x+1=0} \atop {x-y=1}} \right.$

$x^{2} -4x+1=0$
ecuatia de gradul II

x²-4x+1=0
Δ=b²-4ac=16- 4=12
√Δ=√3×2²=2√3
x₁/₂=(-b+-√Δ)/2a=(4+-2√3)/2=2+-√3

x₁=(4+√12)/2=(4+2√3)/2=2+√3
x₂=(4-√12)/2=(4-2√3)/2=2-√3

dar y =x-1
deci  y₁/₂=x₁/₂-1
y₁=x₁-1⇒ y₁=2+√3-1=1+√3
y₂=x₂-1=2-√3-1=1-√3

deci solutiile vor fi (x₁ ; y₁) adica (2+√3 ; 1+√3)

si (x₂ ; y₂) adica (2-√3 ; 1+-3)