Question

(х-1)^2< корень 2(х-1)

Answer 1

(x-1)²<√2(x-1)
x≥1
x²-2x+1<√2x-2
√2x-2>x²-2x+1
-2x+1≥0
-2x+1<0
x∈(1; ∛2 +1) - ответ
х≥1

Answer 2

To solve the inequality $(x-1)^2 < \sqrt{2(x-1)}$, we'll need to consider different cases for the value of $x$ based on the domain where the inequality holds true. First, let's simplify the inequality:

$(x-1)^2 < \sqrt{2(x-1)}$

Now, let $y = x-1$, then we have:

$y^2 < \sqrt{2y}$

To solve this, we need to consider two cases:

Case 1: $y > 0$ In this case, the square root of $2y$ is positive, so we can square both sides of the inequality without changing the inequality's direction:

$y^2 < 2y$

Now, we can rewrite the inequality in the standard form:

$y^2 - 2y < 0$

Factorizing the left-hand side:

$y(y - 2) < 0$

Now, we have two critical points where the inequality might change: $y = 0$ and $y = 2$. Test the inequality for values in each interval:

For $y < 0$: Pick $y = -1$ (any value less than 0) $(-1)(-1 - 2) < 0 \implies 3 < 0$ (False)

For $0 < y < 2$: Pick $y = 1$ (any value between 0 and 2) $1(1 - 2) < 0 \implies -1 < 0$ (True)

For $y > 2$: Pick $y = 3$ (any value greater than 2) $3(3 - 2) < 0 \implies 3 < 0$ (False)

Thus, the inequality holds true for $0 < y < 2$ (or equivalently, $0 < x - 1 < 2$).

Case 2: $y < 0$ In this case, the square root of $2y$ is not a real number, as $2y$ would be negative, and taking the square root of a negative number would result in an imaginary number. Hence, there are no valid solutions for $y < 0$ (or $x - 1 < 0$).

So, combining the solutions from both cases, we have:

$0 < x - 1 < 2$

Now, add 1 to all parts of the inequality:

$1 < x < 3$

The solution to the inequality is $1 < x < 3$.