Question

Система x²+y²=5 log2x+log2y=1

Answer 1

$\log_{2}x+\log_{2}y=\log_{2}xy=1 \Leftrightarrow xy=2 \Leftrightarrow 2xy=4$
Сложим это уравнение с уравнением  $x^{2}+y^{2}=5$
Получим: 
$(x+y)^{2}=9 \Leftrightarrow x+y= \pm 3$
1) Возьмем сначала  $x+y=3$;  $x=3-y$
Тогда 
$(3-y)y=2 \Leftrightarrow -y^{2}+3y-2=0 \Leftrightarrow y=1,y=2$
И x=2, x=1; Если $x+y=-3$, то получаем уравнение (относительно y)   $-y^{2}-3y-2=0 \Leftrightarrow y=-2,y=-2$ и x=-1,x=-2; Хотя можно было заметить, что перед нами симметрический многочлен второй четной степени и достаточно было рассмотреть один случай, а дальше комбинации со знаками
Ответ: (2;1) (-2;-1) (1;2) (-1;-2)

Answer 2

To solve the system of equations x² + y² = 5 and log₂(x) + log₂(y) = 1, we can use both algebraic and logarithmic properties. Let's go step by step:

Step 1: Simplify the second equation using logarithmic properties. Since log₂(x) + log₂(y) = 1, we can combine the logarithms using the product rule of logarithms:

log₂(xy) = 1

Step 2: Rewrite the equation in exponential form. Since log₂(xy) = 1, we can rewrite it as an exponential equation:

2^1 = xy

Step 3: Simplify the exponential equation.

xy = 2

Step 4: Now we have a system of equations:

1. x² + y² = 5
2. xy = 2

Step 5: Solve the system of equations. We can use substitution or elimination method here.

Method 1 (Substitution): From equation 2, we can express y in terms of x:

y = 2/x

Now substitute this value of y into equation 1:

x² + (2/x)² = 5

Simplify the equation:

x² + 4/x² = 5

Multiply the entire equation by x² to get rid of the fraction:

x⁴ + 4 = 5x²

Now rearrange the equation:

x⁴ - 5x² + 4 = 0

This is a quadratic equation in terms of x². We can factor it:

(x² - 1)(x² - 4) = 0

Now set each factor to zero and solve for x:

1. x² - 1 = 0 x² = 1 x = ±1

2. x² - 4 = 0 x² = 4 x = ±2

Now that we have the values of x, we can find the corresponding values of y using equation 2:

For x = 1, y = 2/x = 2/1 = 2 For x = -1, y = 2/x = 2/-1 = -2 For x = 2, y = 2/x = 2/2 = 1 For x = -2, y = 2/x = 2/-2 = -1

So, the solutions to the system of equations are: (x, y) = (1, 2), (-1, -2), (2, 1), (-2, -1)

Method 2 (Elimination): From equation 2, we know that xy = 2. Now, we can express y in terms of x:

y = 2/x

Substitute this value of y into equation 1:

x² + (2/x)² = 5

Simplify the equation:

x² + 4/x² = 5

Now, multiply the entire equation by x² to get rid of the fraction:

x⁴ + 4 = 5x²

Rearrange the equation:

x⁴ - 5x² + 4 = 0

This is a quadratic equation in terms of x². We can factor it:

(x² - 1)(x² - 4) = 0

Now set each factor to zero and solve for x:

1. x² - 1 = 0 x² = 1 x = ±1

2. x² - 4 = 0 x² = 4 x = ±2

Now that we have the values of x, we can find the corresponding values of y using equation 2:

For x = 1, y = 2/x = 2/1 = 2 For x = -1, y = 2/x = 2/-1 = -2 For x = 2, y = 2/x = 2/2 = 1 For x = -2, y = 2/x = 2/-2 = -1

So, the solutions to the system of equations are: (x, y) = (1, 2), (-1, -2), (2, 1), (-2, -1)

Both methods yield the same solutions for the system of equations.