Question

1)4(х-3)<3(х+1)+х 2)(х-3)піднести до квадрату> -х(6-х)

Answer 1

1) 4х-12<3х+3+х
4х-12<4х+3
-12<3

Answer 2

To solve the inequalities, let's work through each of them step by step:

1. $4(x-3) < 3(x+1) + x$

First, we will distribute the terms inside the parentheses:

$4x - 12 < 3x + 3 + x$

Now, combine like terms:

$4x - 12 < 4x + 3$

Next, let's isolate the variable $x$ on one side of the inequality:

Subtract $4x$ from both sides:

$-12 < 3$

The inequality $-12 < 3$ is true regardless of the value of $x$. This means that the original inequality $4(x-3) < 3(x+1) + x$ is true for all real values of $x$.

So, the solution to the inequality is $-\infty < x < \infty$, which means $x$ can be any real number.

2. $(x-3)^2 > -x(6-x)$

First, simplify both sides of the inequality:

$(x-3)^2 > -x(6-x)$

Simplify the left side by squaring the binomial:

$x^2 - 6x + 9 > -x(6-x)$

Expand the right side:

$x^2 - 6x + 9 > -6x + x^2$

Now, let's get all the terms on one side of the inequality:

$x^2 - 6x + 9 + 6x - x^2 > 0$

Simplify and combine like terms:

$-6x + 9 > 0$

Next, isolate the variable $x$ on one side:

Subtract 9 from both sides:

$-6x > -9$

Now, divide both sides by -6, remembering to reverse the inequality when dividing by a negative number:

$x < \frac{9}{6}$

$x < \frac{3}{2}$

So, the solution to the inequality is $x < \frac{3}{2}$.

Putting it all together, the solutions to the two inequalities are:

1. $-\infty < x < \infty$

2. $x < \frac{3}{2}$