Question

(lg^2x -3lgx+3)/(lgx-1) ≤ 1

Answer 1

Введём замену lgx = t.

Единицу перенесём влево и приведём к общему знаменателю.

((t² - 3t + 3 - t + 1)/(t - 1)) ≤ 0

((t² - 4t + 4)/(t - 1)) ≤ 0   ОДЗ: t ≠ 1.

((t - 2)²*(t - 1) ≤ 0

Имеем 1 точку, в которых функция превращается в ноль: t = 2.

Также имеем 1 промежуток от 0 до 1, отвечающий условию задания (по ОДЗ).

Обратная замена:  lgx = 2, х = 10² = 100.

                                 lgx = 0...1,   0 < x < 10.

Ответ:  0 < x < 10,  х = 100.

 

Answer 2

To solve the inequality $\frac{{\lg^2x - 3\lg x + 3}}{{\lg x - 1}} \leq 1$, we need to consider two cases:

1. Case: $\lg x - 1 > 0$
2. Case: $\lg x - 1 < 0$

Let's start with the first case:

1. Case: $\lg x - 1 > 0$

In this case, $\lg x > 1$, and we can rewrite the inequality as follows:

$\lg^2x - 3\lg x + 3 \leq \lg x - 1$

Now, let's solve this quadratic inequality. Move all terms to the left-hand side:

$\lg^2x - 3\lg x + 3 - \lg x + 1 \leq 0$

Simplify:

$\lg^2x - 4\lg x + 4 \leq 0$

Now, we can factor this expression:

$(\lg x - 2)^2 \leq 0$

Since the square of any real number is non-negative, the only way for this inequality to hold is when $\lg x - 2 = 0$.

Solve for $\lg x$:

$\lg x = 2$

So, in this case, $x = 10^2 = 100$.

Now, let's move on to the second case:

2. Case: $\lg x - 1 < 0$

In this case, $\lg x < 1$, and we can rewrite the inequality as follows:

$\lg^2x - 3\lg x + 3 \geq \lg x - 1$

Now, let's solve this quadratic inequality. Move all terms to the left-hand side:

$\lg^2x - 3\lg x + 3 - \lg x + 1 \geq 0$

Simplify:

$\lg^2x - 4\lg x + 4 \geq 0$

Now, we can factor this expression:

$(\lg x - 2)^2 \geq 0$

As we mentioned earlier, the square of any real number is non-negative, which means this inequality holds true for all values of $\lg x$.

However, we have the restriction that $\lg x < 1$ for this case, which means $x$ must be in the interval $0 < x < 10$. But, we also have the previous solution from the first case, $x = 100$, which does not satisfy $\lg x < 1$.

Therefore, the only solution that satisfies the original inequality is $x = 100$.