Let's solve each of the options step by step:
Option #1: \(3(2x+5) = 6x + 15\)
Distributing on the left side: \[ 3 \cdot 2x + 3 \cdot 5 = 6x + 15 \] \[ 6x + 15 = 6x + 15 \]
This equation is true for all values of \(x\). Thus, Option #1 is an identity.
Option #2: \(8x+7=5x+3x+1+6\)
First, simplify the right side: \[ 8x + 7 = (5x + 3x) + 1 + 6 \] \[ 8x + 7 = 8x + 7 \]
This equation is also true for all values of \(x\). Thus, Option #2 is an identity.
Option #3: \((2x^2 − 4)(3y + 6) = 6x^2y + 12x^2 − 12y − 24\)
First, let's check the left-hand side. \[ (2x^2 - 4)(3y + 6) \]
Distributing: \[ 2x^2 \cdot 3y + 2x^2 \cdot 6 - 4 \cdot 3y - 4 \cdot 6 \] \[ = 6x^2y + 12x^2 - 12y - 24 \]
The left-hand side simplifies to the same as the right-hand side: \[ 6x^2y + 12x^2 - 12y - 24 = 6x^2y + 12x^2 - 12y - 24 \]
This equation holds true, so Option #3 is also an identity.
Option #4: \(2x^{12} \cdot 3 = 6x^{\frac{1}{2}}\)
First, simplify the left-hand side: \[ 2x^{12} \cdot 3 = 6x^{12} \]
Now, we compare it with the right-hand side: \[ 6x^{12} = 6x^{\frac{1}{2}} \]
Dividing both sides by 6 gives: \[ x^{12} = x^{\frac{1}{2}} \]
This only holds true if \(x \neq 0\) and under certain conditions. Specifically: \[ 12 = \frac{1}{2} \Rightarrow 24 = 1 \] which is not true.
Thus, Option #4 does not hold true for all \(x\) values.
Conclusion:
- Option #1: Identity
- Option #2: Identity
- Option #3: Identity
- Option #4: Not valid for all \(x\)
The only option that is not valid universally is Option #4.