Solve (2019+1/11d_{1})frac{sqrt{24}+sqrt{8}}{sqrt{2}}-(sqrt{3})^0=

$$( 2019 + \frac { 1 } { 11 } d _ { 1 } ) \frac { \sqrt { 24 } + \sqrt { 8 } } { \sqrt { 2 } } - ( \sqrt { 3 } ) ^ { 0 } =$$

Factor

$$\frac{\sqrt{2}\left(\left(2\sqrt{2}+2\sqrt{6}\right)\left(d_{1}+22209\right)-11\sqrt{2}\right)}{22}$$

Solution Steps

Factor 24=2^{2}\times 6. Rewrite the square root of the product \sqrt{2^{2}\times 6} as the product of square roots \sqrt{2^{2}}\sqrt{6}. Take the square root of 2^{2}.

$$factor(\left(2019+\frac{1}{11}d_{1}\right)\times \frac{2\sqrt{6}+\sqrt{8}}{\sqrt{2}}-\left(\sqrt{3}\right)^{0})$$

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

$$factor(\left(2019+\frac{1}{11}d_{1}\right)\times \frac{2\sqrt{6}+2\sqrt{2}}{\sqrt{2}}-\left(\sqrt{3}\right)^{0})$$

Rationalize the denominator of \frac{2\sqrt{6}+2\sqrt{2}}{\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

$$factor(\left(2019+\frac{1}{11}d_{1}\right)\times \frac{\left(2\sqrt{6}+2\sqrt{2}\right)\sqrt{2}}{\left(\sqrt{2}\right)^{2}}-\left(\sqrt{3}\right)^{0})$$

The square of \sqrt{2} is 2.

$$factor(\left(2019+\frac{1}{11}d_{1}\right)\times \frac{\left(2\sqrt{6}+2\sqrt{2}\right)\sqrt{2}}{2}-\left(\sqrt{3}\right)^{0})$$

Use the distributive property to multiply 2019+\frac{1}{11}d_{1} by \frac{\left(2\sqrt{6}+2\sqrt{2}\right)\sqrt{2}}{2}.

$$factor(2019\times \frac{\left(2\sqrt{6}+2\sqrt{2}\right)\sqrt{2}}{2}+\frac{1}{11}d_{1}\times \frac{\left(2\sqrt{6}+2\sqrt{2}\right)\sqrt{2}}{2}-\left(\sqrt{3}\right)^{0})$$

Use the distributive property to multiply 2\sqrt{6}+2\sqrt{2} by \sqrt{2}.

$$factor(2019\times \frac{2\sqrt{6}\sqrt{2}+2\left(\sqrt{2}\right)^{2}}{2}+\frac{1}{11}d_{1}\times \frac{\left(2\sqrt{6}+2\sqrt{2}\right)\sqrt{2}}{2}-\left(\sqrt{3}\right)^{0})$$

Factor 6=2\times 3. Rewrite the square root of the product \sqrt{2\times 3} as the product of square roots \sqrt{2}\sqrt{3}.

$$factor(2019\times \frac{2\sqrt{2}\sqrt{3}\sqrt{2}+2\left(\sqrt{2}\right)^{2}}{2}+\frac{1}{11}d_{1}\times \frac{\left(2\sqrt{6}+2\sqrt{2}\right)\sqrt{2}}{2}-\left(\sqrt{3}\right)^{0})$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$factor(2019\times \frac{2\times 2\sqrt{3}+2\left(\sqrt{2}\right)^{2}}{2}+\frac{1}{11}d_{1}\times \frac{\left(2\sqrt{6}+2\sqrt{2}\right)\sqrt{2}}{2}-\left(\sqrt{3}\right)^{0})$$

Multiply 2 and 2 to get 4.

$$factor(2019\times \frac{4\sqrt{3}+2\left(\sqrt{2}\right)^{2}}{2}+\frac{1}{11}d_{1}\times \frac{\left(2\sqrt{6}+2\sqrt{2}\right)\sqrt{2}}{2}-\left(\sqrt{3}\right)^{0})$$

The square of \sqrt{2} is 2.

$$factor(2019\times \frac{4\sqrt{3}+2\times 2}{2}+\frac{1}{11}d_{1}\times \frac{\left(2\sqrt{6}+2\sqrt{2}\right)\sqrt{2}}{2}-\left(\sqrt{3}\right)^{0})$$

Multiply 2 and 2 to get 4.

$$factor(2019\times \frac{4\sqrt{3}+4}{2}+\frac{1}{11}d_{1}\times \frac{\left(2\sqrt{6}+2\sqrt{2}\right)\sqrt{2}}{2}-\left(\sqrt{3}\right)^{0})$$

Express 2019\times \frac{4\sqrt{3}+4}{2} as a single fraction.

$$factor(\frac{2019\left(4\sqrt{3}+4\right)}{2}+\frac{1}{11}d_{1}\times \frac{\left(2\sqrt{6}+2\sqrt{2}\right)\sqrt{2}}{2}-\left(\sqrt{3}\right)^{0})$$

Use the distributive property to multiply 2\sqrt{6}+2\sqrt{2} by \sqrt{2}.

$$factor(\frac{2019\left(4\sqrt{3}+4\right)}{2}+\frac{1}{11}d_{1}\times \frac{2\sqrt{6}\sqrt{2}+2\left(\sqrt{2}\right)^{2}}{2}-\left(\sqrt{3}\right)^{0})$$

Factor 6=2\times 3. Rewrite the square root of the product \sqrt{2\times 3} as the product of square roots \sqrt{2}\sqrt{3}.

$$factor(\frac{2019\left(4\sqrt{3}+4\right)}{2}+\frac{1}{11}d_{1}\times \frac{2\sqrt{2}\sqrt{3}\sqrt{2}+2\left(\sqrt{2}\right)^{2}}{2}-\left(\sqrt{3}\right)^{0})$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$factor(\frac{2019\left(4\sqrt{3}+4\right)}{2}+\frac{1}{11}d_{1}\times \frac{2\times 2\sqrt{3}+2\left(\sqrt{2}\right)^{2}}{2}-\left(\sqrt{3}\right)^{0})$$

Multiply 2 and 2 to get 4.

$$factor(\frac{2019\left(4\sqrt{3}+4\right)}{2}+\frac{1}{11}d_{1}\times \frac{4\sqrt{3}+2\left(\sqrt{2}\right)^{2}}{2}-\left(\sqrt{3}\right)^{0})$$

The square of \sqrt{2} is 2.

$$factor(\frac{2019\left(4\sqrt{3}+4\right)}{2}+\frac{1}{11}d_{1}\times \frac{4\sqrt{3}+2\times 2}{2}-\left(\sqrt{3}\right)^{0})$$

Multiply 2 and 2 to get 4.

$$factor(\frac{2019\left(4\sqrt{3}+4\right)}{2}+\frac{1}{11}d_{1}\times \frac{4\sqrt{3}+4}{2}-\left(\sqrt{3}\right)^{0})$$

Multiply \frac{1}{11} times \frac{4\sqrt{3}+4}{2} by multiplying numerator times numerator and denominator times denominator.

$$factor(\frac{2019\left(4\sqrt{3}+4\right)}{2}+\frac{4\sqrt{3}+4}{11\times 2}d_{1}-\left(\sqrt{3}\right)^{0})$$

Use the distributive property to multiply 2019 by 4\sqrt{3}+4.

$$factor(\frac{8076\sqrt{3}+8076}{2}+\frac{4\sqrt{3}+4}{11\times 2}d_{1}-\left(\sqrt{3}\right)^{0})$$

Multiply 11 and 2 to get 22.

$$factor(\frac{8076\sqrt{3}+8076}{2}+\frac{4\sqrt{3}+4}{22}d_{1}-\left(\sqrt{3}\right)^{0})$$

Express \frac{4\sqrt{3}+4}{22}d_{1} as a single fraction.

$$factor(\frac{8076\sqrt{3}+8076}{2}+\frac{\left(4\sqrt{3}+4\right)d_{1}}{22}-\left(\sqrt{3}\right)^{0})$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2 and 22 is 22. Multiply \frac{8076\sqrt{3}+8076}{2} times \frac{11}{11}.

$$factor(\frac{11\left(8076\sqrt{3}+8076\right)}{22}+\frac{\left(4\sqrt{3}+4\right)d_{1}}{22}-\left(\sqrt{3}\right)^{0})$$

Since \frac{11\left(8076\sqrt{3}+8076\right)}{22} and \frac{\left(4\sqrt{3}+4\right)d_{1}}{22} have the same denominator, add them by adding their numerators.

$$factor(\frac{11\left(8076\sqrt{3}+8076\right)+\left(4\sqrt{3}+4\right)d_{1}}{22}-\left(\sqrt{3}\right)^{0})$$

Do the multiplications in 11\left(8076\sqrt{3}+8076\right)+\left(4\sqrt{3}+4\right)d_{1}.

$$factor(\frac{88836\sqrt{3}+88836+4\sqrt{3}d_{1}+4d_{1}}{22}-\left(\sqrt{3}\right)^{0})$$

Calculate \sqrt{3} to the power of 0 and get 1.

$$factor(\frac{88836\sqrt{3}+88836+4\sqrt{3}d_{1}+4d_{1}}{22}-1)$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{22}{22}.

$$factor(\frac{88836\sqrt{3}+88836+4\sqrt{3}d_{1}+4d_{1}}{22}-\frac{22}{22})$$

Since \frac{88836\sqrt{3}+88836+4\sqrt{3}d_{1}+4d_{1}}{22} and \frac{22}{22} have the same denominator, subtract them by subtracting their numerators.

$$factor(\frac{88836\sqrt{3}+88836+4\sqrt{3}d_{1}+4d_{1}-22}{22})$$

Combine like terms in 88836\sqrt{3}+88836+4\sqrt{3}d_{1}+4d_{1}-22.

$$factor(\frac{88836\sqrt{3}+88814+4\sqrt{3}d_{1}+4d_{1}}{22})$$

Consider 88836\times 3^{\frac{1}{2}}+88814+4\times 3^{\frac{1}{2}}d_{1}+4d_{1}. Factor out 2.

$$2\left(44418\sqrt{3}+44407+2\sqrt{3}d_{1}+2d_{1}\right)$$

Rewrite the complete factored expression. Simplify.

$$\frac{44418\sqrt{3}+44407+2\sqrt{3}d_{1}+2d_{1}}{11}$$

Evaluate

$$\frac{\sqrt{2}\left(\sqrt{2}+\sqrt{6}\right)\left(d_{1}+22209\right)-11}{11}$$