Question

(x-2)(x+1)-1 ≤(x-1)²

Answer 1

Ответ:

x^2 - 2x + x - 2 - 1 <= x^2 - 2x + 1

-x - 3 <= -2x + 1

x <= 4

Пошаговое объяснение:

Answer 2

Ответ:

(x-2)(x+1)-1 ≤(x-1)²

$x * x-2*x+x-2*1 -1 \leq x^{2} -2x+1;\\\\x^{2} +x-2x-2-1\leq x^{2} -2x+1;\\\\x-2x-2-1\leq -2x+1;\\\\x-2-1\leq 1;\\\\x-3\leq 1;\\\\x\leq 1+3;\\\\x\leq 4.$

Во второй строке в обеих сторонах сокращаются  $x^{2}$, от них останется 0.

В третьей строке в обеих сторонах сокращаются  $-2x$, от них останется 0.

Answer 3

To solve the inequality $(x-2)(x+1) - 1 \leq (x-1)^2$, you can follow these steps:

1. Expand both sides of the inequality: $(x-2)(x+1) - 1 \leq (x-1)^2$ becomes $(x^2 - x - 2) - 1 \leq x^2 - 2x + 1$.

2. Simplify both sides of the inequality: $x^2 - x - 2 - 1 \leq x^2 - 2x + 1$.

3. Combine like terms: $x^2 - x - 3 \leq x^2 - 2x + 1$.

4. Subtract $x^2$ from both sides to cancel out the quadratic terms: $-x - 3 \leq -2x + 1$.

5. Add $2x$ to both sides to isolate the variable terms on one side: $-x + 2x - 3 \leq 1$.

6. Combine like terms: $x - 3 \leq 1$.

7. Add 3 to both sides to isolate the x-term: $x \leq 1 + 3$.

8. Simplify: $x \leq 4$.

So, the solution to the inequality is $x \leq 4$.