Question

1) 22 − 19 − 10 ≤ 0; 2) (4х-9)(1-х) > 0;3) х2-8х+16 ≥0.

Answer 1

Ответ:

1)22-19-10≤0

-7≤0

2)(4x-9)(1-x)>0

4x-4x²-9+9x>0

-4x²+13x-9>0      |*(-1)

4x²-13x+9=0

D=169-4*4*9=169-144=25=5²

x1=(13+5)/(2*4)=18/8=9/4

x2=(13-5)/(2*4)=8/8=1

3)x²-8x+16≥0

x²-8x+16=0

D=64-4*16=64-64=0

x=8/2=4

Объяснение:

Answer 2

Let's solve each of the given inequalities:

1. $22 - 19 - 10 \leq 0$: $22 - 19 - 10 = -7$, and $-7$ is less than or equal to $0$, so this inequality is true.

2. $(4x - 9)(1 - x) > 0$: To solve this inequality, we'll first find the critical points where the expression equals zero, i.e., $4x - 9 = 0$ and $1 - x = 0$: $4x - 9 = 0 \Rightarrow x = \frac{9}{4}$ and $1 - x = 0 \Rightarrow x = 1$.

   We can then use these critical points to divide the real number line into intervals and test the expression's sign in each interval:

   • For $x < \frac{9}{4}$, both factors are negative, so the expression is positive.
   • For $\frac{9}{4} < x < 1$, $4x - 9$ is positive and $1 - x$ is negative, so the expression is negative.
   • For $x > 1$, both factors are positive, so the expression is positive.

   Therefore, the solution to $(4x - 9)(1 - x) > 0$ is $x < \frac{9}{4}$ or $x > 1$.

3. $x^2 - 8x + 16 \geq 0$: To solve this inequality, we'll find the critical points where the expression equals zero, i.e., $x^2 - 8x + 16 = 0$. We can factor this quadratic equation: $x^2 - 8x + 16 = (x - 4)^2 = 0$.

   The critical point is $x = 4$. Now, we'll use this critical point to divide the real number line into intervals and test the expression's sign in each interval:

   • For $x < 4$, $(x - 4)^2$ is positive, so the expression is positive.
   • For $x > 4$, $(x - 4)^2$ is positive, so the expression is positive.
   • At $x = 4$, $(x - 4)^2 = 0$, so the expression is equal to $0$.

   Therefore, the solution to $x^2 - 8x + 16 \geq 0$ is $x \leq 4$.