Solve |-5/2+1|/sqrt{5^2+1^2+1^2}

$$\frac { | - \frac { 5 } { 2 } + 1 | } { \sqrt { 5 ^ { 2 } + 1 ^ { 2 } + 1 ^ { 2 } } }$$

Evaluate

$$\frac{\sqrt{3}}{6}\approx 0.288675135$$

Solution Steps

Convert 1 to fraction \frac{2}{2}.

$$\frac{|-\frac{5}{2}+\frac{2}{2}|}{\sqrt{5^{2}+1^{2}+1^{2}}}$$

Since -\frac{5}{2} and \frac{2}{2} have the same denominator, add them by adding their numerators.

$$\frac{|\frac{-5+2}{2}|}{\sqrt{5^{2}+1^{2}+1^{2}}}$$

Add -5 and 2 to get -3.

$$\frac{|-\frac{3}{2}|}{\sqrt{5^{2}+1^{2}+1^{2}}}$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of -\frac{3}{2} is \frac{3}{2}.

$$\frac{\frac{3}{2}}{\sqrt{5^{2}+1^{2}+1^{2}}}$$

Calculate 5 to the power of 2 and get 25.

$$\frac{\frac{3}{2}}{\sqrt{25+1^{2}+1^{2}}}$$

Calculate 1 to the power of 2 and get 1.

$$\frac{\frac{3}{2}}{\sqrt{25+1+1^{2}}}$$

Add 25 and 1 to get 26.

$$\frac{\frac{3}{2}}{\sqrt{26+1^{2}}}$$

Calculate 1 to the power of 2 and get 1.

$$\frac{\frac{3}{2}}{\sqrt{26+1}}$$

Add 26 and 1 to get 27.

$$\frac{\frac{3}{2}}{\sqrt{27}}$$

Factor 27=3^{2}\times 3. Rewrite the square root of the product \sqrt{3^{2}\times 3} as the product of square roots \sqrt{3^{2}}\sqrt{3}. Take the square root of 3^{2}.

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

Express \frac{\frac{3}{2}}{3\sqrt{3}} as a single fraction.

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

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

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

The square of \sqrt{3} is 3.

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

Cancel out 3 in both numerator and denominator.

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

Multiply 2 and 3 to get 6.

$$\frac{\sqrt{3}}{6}$$