№1 а) 5х-3>4 б) 4(х-4)≤12 №2 {0,2(х-5)<2 {0,4(х+1)≥-1 №3 (х+2)^2(х-3)/х+1 ≤0 №4 |2х-4|<7 №5 а) 4х+2<7 б) 3(х+2)≥11 №6 {0,5(х-4)<4 {0,2(х+2)≥-2 №7 (х-4)^2(х+3)/(х-2) ≥0 №8 |3х+1|>4
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15 января 2025 17:35
Let's solve each of the inequalities step by step.
№1
a) ( 5x - 3 > 4 )
- Add 3 to both sides: [ 5x > 7 ]
- Divide by 5: [ x > \frac{7}{5} ] [ x > 1.4 ]
b) ( 4(x - 4) \leq 12 )
- Distribute 4: [ 4x - 16 \leq 12 ]
- Add 16 to both sides: [ 4x \leq 28 ]
- Divide by 4: [ x \leq 7 ]
Summary for №1:
- a) ( x > 1.4 )
- b) ( x \leq 7 )
№2
a) ( 0.2(x - 5) < 2 )
- Multiply both sides by 10 (to eliminate the decimal): [ 2(x - 5) < 20 ]
- Distribute: [ 2x - 10 < 20 ]
- Add 10 to both sides: [ 2x < 30 ]
- Divide by 2: [ x < 15 ]
b) ( 0.4(x + 1) \geq -1 )
- Multiply both sides by 10: [ 4(x + 1) \geq -10 ]
- Distribute: [ 4x + 4 \geq -10 ]
- Subtract 4 from both sides: [ 4x \geq -14 ]
- Divide by 4: [ x \geq -\frac{14}{4} ] [ x \geq -3.5 ]
Summary for №2:
- a) ( x < 15 )
- b) ( x \geq -3.5 )
№3
[ \frac{(x+2)^2(x-3)}{x+1} \leq 0 ]
Identify critical points:
- ( (x + 2)^2 = 0 ) ⇒ ( x = -2 ) (double root)
- ( x - 3 = 0 ) ⇒ ( x = 3 )
- ( x + 1 = 0 ) ⇒ ( x = -1 ) (undefined)
Test intervals:
- For ( x < -2 ): Choose ( x = -3 ) → positive
- For ( -2 < x < -1 ): Choose ( x = -1.5 ) → negative
- For ( -1 < x < 3 ): Choose ( x = 0 ) → positive
- For ( x > 3 ): Choose ( x = 4 ) → positive
Summary of intervals:
- Negative or zero on: ( -2 \leq x < -1 ) (include -2, not -1)
Summary for №3:
[ x \in [-2, -1) ]
№4
[ |2x - 4| < 7 ]
This results in two inequalities:
( 2x - 4 < 7 ) [ 2x < 11 ] [ x < 5.5 ]
( 2x - 4 > -7 ) [ 2x > -3 ] [ x > -\frac{3}{2} ] [ x > -1.5 ]
Summary for №4:
[ -1.5 < x < 5.5 ]
№5
a) ( 4x + 2 < 7 )
- Subtract 2 from both sides: [ 4x < 5 ]
- Divide by 4: [ x < \frac{5}{4} ] [ x < 1.25 ]
b) ( 3(x + 2) \geq 11 )
- Distribute: [ 3x + 6 \geq 11 ]
- Subtract 6 from both sides: [ 3x \geq 5 ]
- Divide by 3: [ x \geq \frac{5}{3} ] [ x \geq 1.6667 ]
Summary for №5:
- a) ( x < 1.25 )
- b) ( x \geq 1.6667 )
№6
a) ( 0.5(x - 4) < 4 )
- Multiply both sides by 2: [ x - 4 < 8 ]
- Add 4 to both sides: [ x < 12 ]
b) ( 0.2(x + 2) \geq -2 )
- Multiply both sides by 5: [ x + 2 \geq -10 ]
- Subtract 2 from both sides: [ x \geq -12 ]
Summary for №6:
- a) ( x < 12 )
- b) ( x \geq -12 )
№7
[ \frac{(x-4)^2(x+3)}{x-2} \geq 0 ]
Critical points:
- From ( (x-4)^2 = 0 ): ( x = 4 ) (double root)
- From ( x + 3 = 0 ): ( x = -3 )
- From ( x - 2 = 0 ): ( x = 2 ) (undefined)
Test intervals:
- For ( x < -3 ): Choose ( x = -4 ) → positive
- For ( -3 < x < 2 ): Choose ( x = 0 ) → negative
- For ( 2 < x < 4 ): Choose ( x = 3 ) → positive
- For ( x > 4 ): Choose ( x = 5 ) → positive
Summary for №7:
[ x \in (-\infty, -3) \cup \left(2, 4\right] ]
№8
[ |3x + 1| > 4 ]
This results in two inequalities:
( 3x + 1 > 4 ) [ 3x > 3 ] [ x > 1 ]
( 3x + 1 < -4 ) [ 3x < -5 ] [ x < -\frac{5}{3} ]
Summary for №8:
[ x < -\frac{5}{3} \text{ or } x > 1 ]
You can now summarize these results for your exam preparation! If you have any specific questions or need further clarifications, feel free to ask!
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