Question

(х-4)^2*(x^2-2x-3)<=0 <= - меньше или равно

Answer 1

(x-4):2*(x)=0-986-меньше целого которое ровно одному или нулю

Answer 2

(x-4)² *(x²-2*x-3)≤0
(x-4)²=0 или (x²-2*x-3)=0
(x-4)*(x-4)=0  
x-4=0 
x=4
x²-2*x-3 =0
D=4-4*1*(-3)=4+12=16
x1 = -1 x2=3
указанная функция меньше или равна 0 на интералах
(-∞;-1]∪[0;3]

Answer 3

To solve the inequality $(x-4)^2 \cdot (x^2-2x-3) \leq 0$, we need to find the values of $x$ that satisfy this inequality. To do that, we'll follow these steps:

Step 1: Factorize the expression inside the inequality.

Step 2: Identify the critical points where the expression equals zero.

Step 3: Test intervals between critical points to determine the sign of the expression in each interval.

Step 4: Determine the intervals that satisfy the inequality $(x-4)^2 \cdot (x^2-2x-3) \leq 0$.

Let's go through each step:

Step 1: Factorize the expression inside the inequality. $(x-4)^2 \cdot (x^2-2x-3) = (x-4)^2 \cdot (x-3)(x+1)$

Step 2: Identify the critical points where the expression equals zero. The critical points are the values of $x$ that make any of the factors equal to zero. $x-4 = 0$ gives $x = 4$ $x-3 = 0$ gives $x = 3$ $x+1 = 0$ gives $x = -1$

Step 3: Test intervals between critical points to determine the sign of the expression in each interval.

Test $x < -1$: Choose $x = -2$ $(x-4)^2 \cdot (x-3)(x+1) = (-2-4)^2 \cdot (-2-3)(-2+1) = 36 \cdot (-5) \cdot (-1) = 180$, which is positive.

Test $-1 < x < 3$: Choose $x = 0$ $(x-4)^2 \cdot (x-3)(x+1) = (0-4)^2 \cdot (0-3)(0+1) = 16 \cdot (-3) \cdot 1 = -48$, which is negative.

Test $3 < x < 4$: Choose $x = 3.5$ $(x-4)^2 \cdot (x-3)(x+1) = (3.5-4)^2 \cdot (3.5-3)(3.5+1) = 0.25 \cdot 0.5 \cdot 4.5 = 0.5625$, which is positive.

Test $x > 4$: Choose $x = 5$ $(x-4)^2 \cdot (x-3)(x+1) = (5-4)^2 \cdot (5-3)(5+1) = 1 \cdot 2 \cdot 6 = 12$, which is positive.

Step 4: Determine the intervals that satisfy the inequality $(x-4)^2 \cdot (x^2-2x-3) \leq 0$.

The intervals that satisfy the inequality are $-1 \leq x \leq 3$.

So, the solution to the inequality is $x \in [-1, 3]$.