Solve P=2/(sqrt{2sqrt{3-3+1})^2}

$$P = \frac { 2 } { ( \sqrt { 2 \sqrt { 3 } - 3 + 1 } ) ^ { 2 } }$$

Solve for P

$$P = \frac{\sqrt{3} + 1}{2} \approx 1.366025404$$

Steps for Solving Linear Equation

Add -3 and 1 to get -2.

$$P=\frac{2}{\left(\sqrt{2\sqrt{3}-2}\right)^{2}}$$

Calculate \sqrt{2\sqrt{3}-2} to the power of 2 and get 2\sqrt{3}-2.

$$P=\frac{2}{2\sqrt{3}-2}$$

Rationalize the denominator of \frac{2}{2\sqrt{3}-2} by multiplying numerator and denominator by 2\sqrt{3}+2.

$$P=\frac{2\left(2\sqrt{3}+2\right)}{\left(2\sqrt{3}-2\right)\left(2\sqrt{3}+2\right)}$$

Consider \left(2\sqrt{3}-2\right)\left(2\sqrt{3}+2\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$P=\frac{2\left(2\sqrt{3}+2\right)}{\left(2\sqrt{3}\right)^{2}-2^{2}}$$

Expand \left(2\sqrt{3}\right)^{2}.

$$P=\frac{2\left(2\sqrt{3}+2\right)}{2^{2}\left(\sqrt{3}\right)^{2}-2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$P=\frac{2\left(2\sqrt{3}+2\right)}{4\left(\sqrt{3}\right)^{2}-2^{2}}$$

The square of \sqrt{3} is 3.

$$P=\frac{2\left(2\sqrt{3}+2\right)}{4\times 3-2^{2}}$$

Multiply 4 and 3 to get 12.

$$P=\frac{2\left(2\sqrt{3}+2\right)}{12-2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$P=\frac{2\left(2\sqrt{3}+2\right)}{12-4}$$

Subtract 4 from 12 to get 8.

$$P=\frac{2\left(2\sqrt{3}+2\right)}{8}$$

Divide 2\left(2\sqrt{3}+2\right) by 8 to get \frac{1}{4}\left(2\sqrt{3}+2\right).

$$P=\frac{1}{4}\left(2\sqrt{3}+2\right)$$

Use the distributive property to multiply \frac{1}{4} by 2\sqrt{3}+2.

$$P=\frac{1}{2}\sqrt{3}+\frac{1}{2}$$

Assign P

$$P≔\frac{\sqrt{3}+1}{2}$$