Question

(х-1)*корень из х^2-5x+6 меньше нуля

Answer 1

(х-1)*корень из х^2-5x+6 меньше нуля

корень неотрицателен , поэтому неравенство равносильно системе

 

x-1<0

x^2-5x+6>0

 

x<1

(x-2)(x-3)>0

 

x<1

x<2 или х>3

 

ответ: (-бесконечность; 1)

 

Answer 2

To solve the inequality $(x-1)\sqrt{x^2-5x+6} < 0$, we need to find the values of $x$ that make the expression on the left side less than zero.

First, let's find the critical points by setting each factor equal to zero:

$x-1 = 0$ or $\sqrt{x^2-5x+6} = 0$

From the first equation, we have $x = 1$.

For the second equation, we square both sides to eliminate the square root:

$x^2 - 5x + 6 = 0$

Factoring this quadratic equation, we get:

$(x - 2)(x - 3) = 0$

So, $x = 2$ or $x = 3$.

Now we have three critical points: $x = 1$, $x = 2$, and $x = 3$.

To determine the sign of $(x-1)\sqrt{x^2-5x+6}$ in each interval, we can choose test points within those intervals and evaluate the expression.

1. For $x < 1$, we can choose $x = 0$: $(0 - 1)\sqrt{0^2 - 5(0) + 6} = (-1)\sqrt{6} < 0$ (negative)

2. For $1 < x < 2$, we can choose $x = \frac{3}{2}$: $(\frac{3}{2} - 1)\sqrt{(\frac{3}{2})^2 - 5(\frac{3}{2}) + 6} = (\frac{1}{2})\sqrt{\frac{1}{4} - \frac{15}{2} + 6} = (\frac{1}{2})\sqrt{\frac{1}{4} - \frac{15}{2} + \frac{48}{4}} = (\frac{1}{2})\sqrt{-\frac{59}{4}}$ (complex)

3. For $2 < x < 3$, we can choose $x = \frac{5}{2}$: $(\frac{5}{2} - 1)\sqrt{(\frac{5}{2})^2 - 5(\frac{5}{2}) + 6} = (\frac{3}{2})\sqrt{\frac{25}{4} - \frac{25}{2} + 6} = (\frac{3}{2})\sqrt{\frac{25}{4} - \frac{50}{4} + \frac{24}{4}} = (\frac{3}{2})\sqrt{-\frac{1}{4}}$