Question

В) (х-1)^2+(х-2)^2<1 г) (х+3)(х-2)>3х+10-(х+2)^2

Answer 1

В) (x-1)^2+(x-2)^2<1
x^2-2x+1+x^2-4x+4<1
x^2-2x+x^2-4x+4<1
2x^2-6+4<1 (разделим обе части уравнения на 2)
x^2-3x+2<0
x^2-x-2x+2<0
x*(x-1)-2*(x-1)<0
(x-1)*(x-2)<0
{x-1<0 x-2>0
{x<1 x>2
{x>1 x<2
x∈ (1;2)

г) (x+3)*(x-2)>3x+10-(x+2)^2
x^2-2x+3x-6>3x+10-(x^2+4x+4)
x^2-2x-6>10-x^2-4x-4
x^2-2x-6>6-x^2-4x
x^2-2x-6-6+x^2+4x>0
2x^2+2x-12>0
x^2+x-6>0
x^2+3x-2x-6>0
x*(x+3)-2*(x+3)>0
(x+3)*(x+2)>0
{x+3>0 x-2>0
{x+3<0 x+2<0
{x>-3 x>2
{ x<-3 x<2
x∈(-∞;-3)u(2;+∞)

Answer 2

Let's solve the inequalities step by step:

а) (х-1)^2 + (х-2)^2 < 1

Step 1: Expand the squares. (х^2 - 2х + 1) + (х^2 - 4х + 4) < 1

Step 2: Combine like terms. 2х^2 - 6х + 5 < 1

Step 3: Move all terms to one side of the inequality. 2х^2 - 6х + 4 < 0

Step 4: Simplify the inequality. х^2 - 3х + 2 < 0

Step 5: Factor the quadratic expression. (х - 2)(х - 1) < 0

Step 6: Find the critical points where the expression equals zero. х - 2 = 0 => х = 2 х - 1 = 0 => х = 1

Step 7: Plot the critical points on a number line and check intervals. Now, we need to check the sign of the expression in the intervals between these critical points: (-∞, 1), (1, 2), and (2, +∞).

For х < 1: Choose х = 0 (test point) -> (0 - 2)(0 - 1) = 2 > 0 The expression is positive in this interval.

For 1 < х < 2: Choose х = 1.5 (test point) -> (1.5 - 2)(1.5 - 1) = -0.25 < 0 The expression is negative in this interval.

For х > 2: Choose х = 3 (test point) -> (3 - 2)(3 - 1) = 2 > 0 The expression is positive in this interval.

Step 8: Determine the solution. The inequality is true when the expression is less than zero, i.e., in the interval (1, 2).

b) (х+3)(х-2) > 3х + 10 - (х+2)^2

Step 1: Expand the squared term. (х + 3)(х - 2) > 3х + 10 - (х^2 + 4х + 4)

Step 2: Distribute and combine like terms. х^2 - 2х + 3х - 6 > 3х + 10 - х^2 - 4х - 4

Step 3: Simplify the inequality. х^2 + x - 6 > -х^2 - x + 6

Step 4: Move all terms to one side of the inequality. 2х^2 + 2x - 12 > 0

Step 5: Simplify the inequality. х^2 + x - 6 > 0

Step 6: Factor the quadratic expression. (х - 2)(х + 3) > 0

Step 7: Plot the critical points on a number line and check intervals. Now, we need to check the sign of the expression in the intervals around the critical points: (-∞, -3), (-3, 2), and (2, +∞).

For х < -3: Choose х = -4 (test point) -> (-4 - 2)(-4 + 3) = 2 > 0 The expression is positive in this interval.

For -3 < х < 2: Choose х = 0 (test point) -> (0 - 2)(0 + 3) = -6 < 0 The expression is negative in this interval.

For х > 2: Choose х = 3 (test point) -> (3 - 2)(3 + 3) = 6 > 0 The expression is positive in this interval.

Step 8: Determine the solution. The inequality is true when the expression is greater than zero, i.e., in the intervals (-∞, -3) and (2, +∞).

Final solutions:

а) The solution to the inequality (х-1)^2 + (х-2)^2 < 1 is 1 < х < 2. b) The solution to the inequality (х+3)(х-2) > 3х + 10 - (х+2)^2 is (-∞, -3) and (2, +∞).