Question

Помогите, пожалуйста 1) (2-x)(8-x)<0 2) (1-x)(7-x)>0 3) (x-2)(5+x)>0

Answer 1

1) 16-2x-8x+x²<0
x²-10x+16=0
D= b²-4ac = 100-64= 36
x1= -b-√D
-------- = 10-6
2a. ------ = 2
2

x2= -b+√D
-------- = 10+6
2a. ------ = 8
2
+ - +
——•——•———›
2 8
(-∞;2)€(8;∞)

2) 7-x-7x+x²>0
x²-8x+7=0
D=b²-4ac= 64-28=36
x1= -b-√D = 8-6 =1
-------- -----
2a. 2

x2= -b+√D
-------- = 8+6
2a. ------ = 7
2
+. - +
———•——•——›
1 7
(-∞;1)€(7;∞)

3)5x+x²-10-2x>0
x²+3x-10=0
D=b²-4ac= 9+40=49
x1= -b-√D
-------- = -3-7
2a. ------ = -5
2
x2= -b+√D
-------- = -3+7
2a. ------ = 2
2
+ - +
———•————•———›
-5 2
(-∞;-5)€(2;∞)

Answer 2

To solve the inequality (2-x)(8-x) < 0, we can use the method of signs. First, we find the critical values of x by setting each factor equal to zero:

2 - x = 0 => x = 2 8 - x = 0 => x = 8

Now we have three intervals to consider: (-∞, 2), (2, 8), and (8, ∞). We pick a test point from each interval and check the sign of the expression (2-x)(8-x):

• If x < 2, then we can pick x = 0 as our test point. Plugging this into the expression, we get (2-0)(8-0) = 16, which is positive.
• If 2 < x < 8, then we can pick x = 5 as our test point. Plugging this into the expression, we get (2-5)(8-5) = -9, which is negative.
• If x > 8, then we can pick x = 10 as our test point. Plugging this into the expression, we get (2-10)(8-10) = 16, which is positive.

From this, we can see that the expression (2-x)(8-x) is negative on the interval (2, 8), since it is negative at x = 5. Therefore, the solution to the inequality is:

2 < x < 8

To solve the inequality (1-x)(7-x) > 0, we can again use the method of signs. Finding the critical values of x by setting each factor equal to zero:

1 - x = 0 => x = 1 7 - x = 0 => x = 7

Now we have three intervals to consider: (-∞, 1), (1, 7), and (7, ∞). We pick a test point from each interval and check the sign of the expression (1-x)(7-x):

• If x < 1, then we can pick x = 0 as our test point. Plugging this into the expression, we get (1-0)(7-0) = 7, which is positive.
• If 1 < x < 7, then we can pick x = 5 as our test point. Plugging this into the expression, we get (1-5)(7-5) = -8, which is negative.
• If x > 7, then we can pick x = 8 as our test point. Plugging this into the expression, we get (1-8)(7-8) = 7, which is positive.

From this, we can see that the expression (1-x)(7-x) is positive on the intervals (-∞, 1) and (7, ∞), since it is positive at x = 0 and x = 8. Therefore, the solution to the inequality is:

x < 1 or x > 7

To solve the inequality (x-2)(5+x) > 0, we once again use the method of signs. Finding the critical values of x by setting each factor equal to zero:

x - 2 = 0 => x = 2 5 + x = 0 => x = -5

Now we have three intervals to consider: (-∞, -5), (-5, 2), and (2, ∞). We pick a test point from each interval and check the sign of the expression (x-2)(5+x):