Question

Решите систему: (х-у)^2+(х-у)-2=0 х^2+у^2=41

Answer 1

x²+y²=41

(x-y)²+(x-y)-2=0

Пусть x-y=t   ⇒

t²+t-2=0      D=9   √D=3

t₁=x-y=1     x=y+1       ⇒

(y+1)²+y²=41

y²+2y+1+y²=41

2y²+2y-40=0  |÷2

y²+y-20=0       D=81

y₁=4         x₁=4+1=5

y₂=-5       x₂=-5+1=-4

t₂=x-y=-2         x=y-2

(y-2)²+y²=41

y²-4y+4+y²=41

2y²-4y-37=0      D=164       √D=2√78

y₃=1+√78/2      x₃=1+(√78/2)-2=√78/2-1

y₄=1-√78/2      x₄=1-(√78/2)-2=-1-√78/2

Ответ:  x₁=5     y₁=4      x₂=-4       y₂=-5      

x₃=√78/2-1         y₃=1+√78/2      x₄=-1-√78/2      y₄=1-√78/2.

Answer 2

Let's simplify the first equation by substituting a variable, let's say z = x - y:

z^2 + z - 2 = 0

We can solve for z by factoring the quadratic equation:

(z + 2)(z - 1) = 0

This gives us two possible values for z: z = -2 or z = 1.

Now, let's use these values of z to solve for x and y:

When z = -2, we have x - y = -2 or x = y - 2.

Substituting this into the second equation, we get:

(y-2)^2 + y^2 = 41

Expanding and simplifying, we get:

2y^2 - 4y - 37 = 0

Using the quadratic formula, we get:

y = (4 ± √156)/4 = 1 ± √39/2

This gives us two possible solutions: y = 1 + √39/2 or y = 1 - √39/2.

Substituting these values into x = y - 2, we get:

x = -1 ± √39/2

So the two solutions are:

(x,y) = (-1 + √39/2, 1 + √39/2) or (x,y) = (-1 - √39/2, 1 - √39/2)

When z = 1, we have x - y = 1 or x = y + 1.

Substituting this into the second equation, we get:

(y+1)^2 + y^2 = 41

Expanding and simplifying, we get:

2y^2 + 2y - 39 = 0

Using the quadratic formula, we get:

y = (-2 ± √172)/4 = -1 ± √43/2

This gives us two possible solutions: y = -1 + √43/2 or y = -1 - √43/2.

Substituting these values into x = y + 1, we get:

x = √43/2 or x = -√43/2

So the two solutions are:

(x,y) = (√43/2, -1 + √43/2) or (x,y) = (-√43/2, -1 - √43/2)

Therefore, the complete solution to the system of equations is:

(x,y) = (-1 + √39/2, 1 + √39/2), (-1 - √39/2, 1 - √39/2), (√43/2, -1 + √43/2), or (-√43/2, -1 - √43/2)