Solve sqrt{2^-1+3^-1+4^-1}

$$\sqrt { 2 ^ { - 1 } + 3 ^ { - 1 } + 4 ^ { - 1 } }$$

Evaluate

$$\frac{\sqrt{39}}{6}\approx 1.040833$$

Solution Steps

Calculate 2 to the power of -1 and get \frac{1}{2}.

$$\sqrt{\frac{1}{2}+3^{-1}+4^{-1}}$$

Calculate 3 to the power of -1 and get \frac{1}{3}.

$$\sqrt{\frac{1}{2}+\frac{1}{3}+4^{-1}}$$

Add \frac{1}{2} and \frac{1}{3} to get \frac{5}{6}.

$$\sqrt{\frac{5}{6}+4^{-1}}$$

Calculate 4 to the power of -1 and get \frac{1}{4}.

$$\sqrt{\frac{5}{6}+\frac{1}{4}}$$

Add \frac{5}{6} and \frac{1}{4} to get \frac{13}{12}.

$$\sqrt{\frac{13}{12}}$$

Rewrite the square root of the division \sqrt{\frac{13}{12}} as the division of square roots \frac{\sqrt{13}}{\sqrt{12}}.

$$\frac{\sqrt{13}}{\sqrt{12}}$$

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

$$\frac{\sqrt{13}}{2\sqrt{3}}$$

Rationalize the denominator of \frac{\sqrt{13}}{2\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$\frac{\sqrt{13}\sqrt{3}}{2\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{\sqrt{13}\sqrt{3}}{2\times 3}$$

To multiply \sqrt{13} and \sqrt{3}, multiply the numbers under the square root.

$$\frac{\sqrt{39}}{2\times 3}$$

Multiply 2 and 3 to get 6.

$$\frac{\sqrt{39}}{6}$$