Solve sqrt{36}-sqrt{24/7}*sqrt{49/6}*sqrt{9/7}+sqrt{49}

$$\sqrt { 36 } - \sqrt { \frac { 24 } { 7 } } \cdot \sqrt { \frac { 49 } { 6 } } \cdot \sqrt { \frac { 9 } { 7 } } + \sqrt { 49 }$$

Evaluate

$$7$$

Solution Steps

Calculate the square root of 36 and get 6.

$$6-\sqrt{\frac{24}{7}}\sqrt{\frac{49}{6}}\sqrt{\frac{9}{7}}+\sqrt{49}$$

To multiply \sqrt{\frac{24}{7}} and \sqrt{\frac{49}{6}}, multiply the numbers under the square root.

$$6-\sqrt{28}\sqrt{\frac{9}{7}}+\sqrt{49}$$

Factor 28=2^{2}\times 7. Rewrite the square root of the product \sqrt{2^{2}\times 7} as the product of square roots \sqrt{2^{2}}\sqrt{7}. Take the square root of 2^{2}.

$$6-2\sqrt{7}\sqrt{\frac{9}{7}}+\sqrt{49}$$

Rewrite the square root of the division \sqrt{\frac{9}{7}} as the division of square roots \frac{\sqrt{9}}{\sqrt{7}}.

$$6-2\sqrt{7}\times \frac{\sqrt{9}}{\sqrt{7}}+\sqrt{49}$$

Calculate the square root of 9 and get 3.

$$6-2\sqrt{7}\times \frac{3}{\sqrt{7}}+\sqrt{49}$$

Rationalize the denominator of \frac{3}{\sqrt{7}} by multiplying numerator and denominator by \sqrt{7}.

$$6-2\sqrt{7}\times \frac{3\sqrt{7}}{\left(\sqrt{7}\right)^{2}}+\sqrt{49}$$

The square of \sqrt{7} is 7.

$$6-2\sqrt{7}\times \frac{3\sqrt{7}}{7}+\sqrt{49}$$

Express 2\times \frac{3\sqrt{7}}{7} as a single fraction.

$$6-\frac{2\times 3\sqrt{7}}{7}\sqrt{7}+\sqrt{49}$$

Calculate the square root of 49 and get 7.

$$6-\frac{2\times 3\sqrt{7}}{7}\sqrt{7}+7$$

Multiply 2 and 3 to get 6.

$$6-\frac{6\sqrt{7}}{7}\sqrt{7}+7$$

Express \frac{6\sqrt{7}}{7}\sqrt{7} as a single fraction.

$$6-\frac{6\sqrt{7}\sqrt{7}}{7}+7$$

Multiply \sqrt{7} and \sqrt{7} to get 7.

$$6-\frac{6\times 7}{7}+7$$

Cancel out 7 and 7.

$$6-6+7$$

Subtract 6 from 6 to get 0.

$$0+7$$

Add 0 and 7 to get 7.

$$7$$

Factor

$$7$$