Question

1) I4X-3I<52) I3X-5I=5-3Х

Answer 1

14х-31   -----14  умножить  на  3=42  42-31=11  меньше  52      

Answer 2

1) i4x-3i<5 To solve this inequality, we need to eliminate the imaginary number. We can do this by multiplying both sides of the inequality by the conjugate of -3i, which is 3i:

i4x - 3i < 5 (4x - 3i)(3i) < 5(3i) -9ix + 12x < 15i

Now we can separate the real and imaginary parts:

-9ix + 12x < 15i (12x) < (15i + 9ix)

Since we're only considering real solutions, we can ignore the imaginary part:

12x < 15i + 9ix

This is the final inequality for the given expression.

2) i3x - 5i = 5 - 3x Let's solve this equation step by step:

i3x - 5i = 5 - 3x

To eliminate the imaginary number, we can multiply both sides of the equation by the conjugate of -5i, which is 5i:

(i3x - 5i)(5i) = (5 - 3x)(5i) -15ix^2 + 25i^2 = 25i - 15xi

Since i^2 is equal to -1, we can replace it in the equation:

-15ix^2 + 25(-1) = 25i - 15xi -15ix^2 - 25 = 25i - 15xi

Let's separate the real and imaginary parts again:

-15ix^2 - 25 = 25i - 15xi -15ix^2 + 15xi = 25i + 25

Now we can factor out an i:

i(-15x^2 + 15x) = 25(i + 1)

If we divide both sides of the equation by (i + 1):

-15x^2 + 15x = 25

Now we can solve for x by moving all terms to one side and factorizing if possible:

-15x^2 + 15x - 25 = 0

This is a quadratic equation, so we can use the quadratic formula:

x = (-b ± √(b^2 - 4ac))/(2a)

In this case, a = -15, b = 15, and c = -25.

x = (-15 ± √(15^2 - 4(-15)(-25)))/(2(-15)) x = (-15 ± √(225 - 1500))/(-30) x = (-15 ± √(-1275))/(-30) x = (-15 ± √(1275)i)/(-30)

So the solutions for this equation are x = (-15 + √(1275)i)/(-30) and x = (-15 - √(1275)i)/(-30).