Question

X(6x+5)(3x-33)=0;И конечно-же: |2x-7|=5

Answer 1

1:х(6х+5)(3х-33)=0
7х+5х•3х-33=0
33х=33
х=33:33
х=1


а второе я точно не знаю (((

Answer 2

To solve the equations, let's take them one at a time:

1. $X(6x+5)(3x-33)=0$

To find the solutions for this equation, we set each factor equal to zero and solve for x:

a) $X = 0$

b) $6x + 5 = 0 \Rightarrow 6x = -5 \Rightarrow x = -\frac{5}{6}$

c) $3x - 33 = 0 \Rightarrow 3x = 33 \Rightarrow x = \frac{33}{3} = 11$

So, the solutions to this equation are: $x = -\frac{5}{6}$, $x = 11$, and $X = 0$.

2. $|2x - 7| = 5$

To solve this equation, we need to consider two cases:

a) $2x - 7 = 5$

Solve for x: $2x = 5 + 7 = 12 \Rightarrow x = \frac{12}{2} = 6$

b) $2x - 7 = -5$

Solve for x: $2x = -5 + 7 = 2 \Rightarrow x = \frac{2}{2} = 1$

Now, we have two potential solutions: $x = 6$ and $x = 1$.

However, we also need to consider the absolute value. Since $|2x - 7| = 5$, the expression $2x - 7$ could also be -5:

c) $2x - 7 = -5$

Solve for x: $2x = -5 + 7 = 2 \Rightarrow x = \frac{2}{2} = 1$

But we have already found this solution in case b.

So, the final solutions to the equation $|2x - 7| = 5$ are $x = 6$ and $x = 1$.

To summarize:

1. Solutions to $X(6x+5)(3x-33)=0$ are $x = -\frac{5}{6}$, $x = 11$, and $X = 0$.

2. Solutions to $|2x - 7| = 5$ are $x = 6$ and $x = 1$.