Question

Log2(x^2-x+2) меньше 2

Answer 1

X²-x+2<2²
x²-x+2-4<0
x²-x-2<0
x²-x-2+0
D=1+8=9
x=(1-3)\2=-1
x=(1+3)\2=2
ОДЗ : х∈(-∞;+∞)
х∈(-1;2)

Answer 2

To solve the inequality $\log_2(x^2 - x + 2) < 2$, we need to find the values of $x$ that satisfy the inequality. Here's the step-by-step solution:

Step 1: Remove the logarithm by rewriting the inequality in exponential form: $2^{(\log_2(x^2 - x + 2))} < 2^2$

Step 2: Simplify the right-hand side: $2^{(\log_2(x^2 - x + 2))} < 4$

Step 3: Since $a^{\log_a(b)} = b$, we can rewrite the inequality as: $x^2 - x + 2 < 4$

Step 4: Move all terms to one side of the inequality: $x^2 - x - 2 < 0$

Step 5: Factor the quadratic expression: $(x - 2)(x + 1) < 0$

Step 6: Determine the sign of the expression for different intervals of $x$:

$$\begin{array}{c|ccccc} x & \cdots & -1 & \cdots & 2 & \cdots \\ \hline (x - 2) & - & - & - & 0 & + \\ (x + 1) & - & 0 & + & + & + \\ (x - 2)(x + 1) & + & 0 & - & 0 & + \end{array}$$

Step 7: Find the intervals where $(x - 2)(x + 1) < 0$ (the expression is negative):

The expression is negative for $x \in (-1, 2)$.

Step 8: Finalize the solution:

The solution to the inequality $\log_2(x^2 - x + 2) < 2$ is $x \in (-1, 2)$.