{x³+x²y=12 x²*(x+y)=12 x²*3=12 |÷3 x²=4 x₁=2 x₂=-2 ⇒
{x+y=3 y=3-x y₁=3-2=1 y₂=3-(-2)=3+2=5.
Ответ: x₁=2 y₁=1 x₂=-2 y₂=5.
{x³ + x²y = 12
{x + y = 3
{y = 3-x
x³ + x²·(3-x) = 12
x³ + 3x² - x³ = 12
3x² = 12
x² = 12 : 3
x² = 4
x₁ = - √4 = - 2
x₂ = √4 = 2
x₁ = - 2 => y₁ = 3-(-2) = 3+2= 5
x₂ = 2 => y₂ = 3-2 = 1
Ответ: (-2; 5); (2; 1)
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Answers & Comments
{x³+x²y=12 x²*(x+y)=12 x²*3=12 |÷3 x²=4 x₁=2 x₂=-2 ⇒
{x+y=3 y=3-x y₁=3-2=1 y₂=3-(-2)=3+2=5.
Ответ: x₁=2 y₁=1 x₂=-2 y₂=5.