Question

решите дча примера на листочке

Answer 1

Решить неравенства:

5. |x²-8x+15|≤|15-x²|

6. 3|x-2|+|5x-4|≤10

Решение:

$\boxed 5 \\ \\ \tt |x^2-8x+15|\leq |15-x^2| \\ (x^2-8x+15)^2 \leq (15-x^2)^2 \\ (x^2-8x+15)^2 - (15-x^2)^2 \leq 0 \\ (x^2-8x+15-15+x^2)(x^2-8x+15+15-x^2) \leq 0 \\ (2x^2-8x)(-8x+30) \leq 0 \\ x(x-4)(-4x+15) \leq 0$

___+___[0]___-___[15/4]___+___[4]___-___

x∈[0; 15/4]U[4; +∞)

$\boxed 6 \\ \\ \tt 3|x-2|_{(1)}+|5x-4|_{(2)} \leq 10 \\ \\ x-2=0 \ \ \Rightarrow \ \ x=2 \\ 5x-4=0 \ \ \Rightarrow \ \ x=0,8 \\ \\ 1) \ \ x<0,8 \ \ (1-, \ 2-) \\ \\ -3x+6-5x+4 \leq 10 \\ -8x \leq 0 \\ x \geq 0 \\ \\ x \in [0; \ 0,8) \\ \\ 2) \ \ x \in [0,8; \ 2) \ \ (1-, \ 2+) \\ \\ -3x+6+5x-4 \leq 10 \\ 2x \leq8 \\ x \leq 4 \\ \\x \in [0,8; \ 2) \\ \\ 3) \ \ x \geq 2 \ \ (1+, \ 2+) \\ \\ 3x-6+5x-4 \leq 10 \\ 8x \leq 20 \\ x \leq 2,5 \\ \\ x \in [2; \ 2,5]$

x∈[0; 2,5]