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Вопрос от +79819634515 Карина 15 января 2025 17:35

№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 )

  1. Add 3 to both sides: [ 5x > 7 ]
  2. Divide by 5: [ x > \frac{7}{5} ] [ x > 1.4 ]

b) ( 4(x - 4) \leq 12 )

  1. Distribute 4: [ 4x - 16 \leq 12 ]
  2. Add 16 to both sides: [ 4x \leq 28 ]
  3. Divide by 4: [ x \leq 7 ]

Summary for №1:

  • a) ( x > 1.4 )
  • b) ( x \leq 7 )

№2

a) ( 0.2(x - 5) < 2 )

  1. Multiply both sides by 10 (to eliminate the decimal): [ 2(x - 5) < 20 ]
  2. Distribute: [ 2x - 10 < 20 ]
  3. Add 10 to both sides: [ 2x < 30 ]
  4. Divide by 2: [ x < 15 ]

b) ( 0.4(x + 1) \geq -1 )

  1. Multiply both sides by 10: [ 4(x + 1) \geq -10 ]
  2. Distribute: [ 4x + 4 \geq -10 ]
  3. Subtract 4 from both sides: [ 4x \geq -14 ]
  4. 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:

  1. ( 2x - 4 < 7 ) [ 2x < 11 ] [ x < 5.5 ]

  2. ( 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 )

  1. Subtract 2 from both sides: [ 4x < 5 ]
  2. Divide by 4: [ x < \frac{5}{4} ] [ x < 1.25 ]

b) ( 3(x + 2) \geq 11 )

  1. Distribute: [ 3x + 6 \geq 11 ]
  2. Subtract 6 from both sides: [ 3x \geq 5 ]
  3. 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 )

  1. Multiply both sides by 2: [ x - 4 < 8 ]
  2. Add 4 to both sides: [ x < 12 ]

b) ( 0.2(x + 2) \geq -2 )

  1. Multiply both sides by 5: [ x + 2 \geq -10 ]
  2. 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:

  1. ( 3x + 1 > 4 ) [ 3x > 3 ] [ x > 1 ]

  2. ( 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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№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
Let's solve each of the inequalities step by step. ### №1 **a)** \( 5x - 3 > 4 \) 1. Add 3 to both sides: \[ 5x > 7 \] 2. Divide by 5: \[ x > \frac{7}{5} \] \[ x > 1.4 \] **b)** \( 4(x - 4) \leq 12 \) 1. Distribute 4: \[ 4x - 16 \leq 12 \] 2. Add 16 to both sides: \[ 4x \leq 28 \] 3. Divide by 4: \[ x \leq 7 \] ### Summary for №1: - **a)** \( x > 1.4 \) - **b)** \( x \leq 7 \) ### №2 **a)** \( 0.2(x - 5) < 2 \) 1. Multiply both sides by 10 (to eliminate the decimal): \[ 2(x - 5) < 20 \] 2. Distribute: \[ 2x - 10 < 20 \] 3. Add 10 to both sides: \[ 2x < 30 \] 4. Divide by 2: \[ x < 15 \] **b)** \( 0.4(x + 1) \geq -1 \) 1. Multiply both sides by 10: \[ 4(x + 1) \geq -10 \] 2. Distribute: \[ 4x + 4 \geq -10 \] 3. Subtract 4 from both sides: \[ 4x \geq -14 \] 4. 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: 1. \( 2x - 4 < 7 \) \[ 2x < 11 \] \[ x < 5.5 \] 2. \( 2x - 4 > -7 \) \[ 2x > -3 \] \[ x > -\frac{3}{2} \] \[ x > -1.5 \] ### Summary for №4: \[ -1.5 < x < 5.5 \] ### №5 **a)** \( 4x + 2 < 7 \) 1. Subtract 2 from both sides: \[ 4x < 5 \] 2. Divide by 4: \[ x < \frac{5}{4} \] \[ x < 1.25 \] **b)** \( 3(x + 2) \geq 11 \) 1. Distribute: \[ 3x + 6 \geq 11 \] 2. Subtract 6 from both sides: \[ 3x \geq 5 \] 3. 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)** \( 0.5(x - 4) < 4 \) 1. Multiply both sides by 2: \[ x - 4 < 8 \] 2. Add 4 to both sides: \[ x < 12 \] **b)** \( 0.2(x + 2) \geq -2 \) 1. Multiply both sides by 5: \[ x + 2 \geq -10 \] 2. 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: 1. \( 3x + 1 > 4 \) \[ 3x > 3 \] \[ x > 1 \] 2. \( 3x + 1 < -4 \) \[ 3x < -5 \] \[ x < -\frac{5}{3} \] ### 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!