Автор - Vika2005122

Решите систему уравнений
у +2х=5
2х-ху= -1

Ответ

Автор - MrSolution

Ответ:

(1; 3), (0.5; 4)

Объяснение:

y + 2x = 5 \ 2x - xy =  - 1 \  \ y + xy = 6 \ y(x + 1) = 6 \  \ y = 5 - 2x \   =  > (5 - 2x)(x + 1) = 6 \ 3x - 2 {x}^{2}  - 1 = 0 \ 2 {x}^{2}  - 3x + 1 = 0 \ 2 {x}^{2}  - 2x - x + 1 = 0 \ 2x(x - 1) - (x - 1) = 0 \ (x - 1)(2x - 1) = 0 \ x =  frac{1}{2}  \ x = 1 \  =  > y = 3 \ y = 4

Ответ

Автор - Grib000089522114811

left { {{y+2x=5} atop {2x-xy=-1}} right. \\y+2x=5\y = 5-2x\\2x-xy=-1\2x-x*(5-2x)=-1\2x-5x+2x^{2} =-1\2x^{2} -3+1=0\\x1=frac{3+sqrt{9-4*2*1} }{2*2} =frac{3+1}{4}=1\\x2=frac{3-sqrt{9-4*2*1} }{2*2} =frac{3-1}{4}=0.5

y1+2x1=5\y1+2=5\y1=3\\y2+2x2=5\y2+1=5\y2=4

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