Solve (71/3)^2-(22/3)^2/(5frac{7{9})^2-(42/9)^2}

$$\frac { ( 7 \frac { 1 } { 3 } ) ^ { 2 } - ( 2 \frac { 2 } { 3 } ) ^ { 2 } } { ( 5 \frac { 7 } { 9 } ) ^ { 2 } - ( 4 \frac { 2 } { 9 } ) ^ { 2 } }$$

Evaluate

$$3$$

Solution Steps

Multiply 7 and 3 to get 21.

$$\frac{\left(\frac{21+1}{3}\right)^{2}-\left(\frac{2\times 3+2}{3}\right)^{2}}{\left(\frac{5\times 9+7}{9}\right)^{2}-\left(\frac{4\times 9+2}{9}\right)^{2}}$$

Add 21 and 1 to get 22.

$$\frac{\left(\frac{22}{3}\right)^{2}-\left(\frac{2\times 3+2}{3}\right)^{2}}{\left(\frac{5\times 9+7}{9}\right)^{2}-\left(\frac{4\times 9+2}{9}\right)^{2}}$$

Calculate \frac{22}{3} to the power of 2 and get \frac{484}{9}.

$$\frac{\frac{484}{9}-\left(\frac{2\times 3+2}{3}\right)^{2}}{\left(\frac{5\times 9+7}{9}\right)^{2}-\left(\frac{4\times 9+2}{9}\right)^{2}}$$

Multiply 2 and 3 to get 6.

$$\frac{\frac{484}{9}-\left(\frac{6+2}{3}\right)^{2}}{\left(\frac{5\times 9+7}{9}\right)^{2}-\left(\frac{4\times 9+2}{9}\right)^{2}}$$

Add 6 and 2 to get 8.

$$\frac{\frac{484}{9}-\left(\frac{8}{3}\right)^{2}}{\left(\frac{5\times 9+7}{9}\right)^{2}-\left(\frac{4\times 9+2}{9}\right)^{2}}$$

Calculate \frac{8}{3} to the power of 2 and get \frac{64}{9}.

$$\frac{\frac{484}{9}-\frac{64}{9}}{\left(\frac{5\times 9+7}{9}\right)^{2}-\left(\frac{4\times 9+2}{9}\right)^{2}}$$

Subtract \frac{64}{9} from \frac{484}{9} to get \frac{140}{3}.

$$\frac{\frac{140}{3}}{\left(\frac{5\times 9+7}{9}\right)^{2}-\left(\frac{4\times 9+2}{9}\right)^{2}}$$

Multiply 5 and 9 to get 45.

$$\frac{\frac{140}{3}}{\left(\frac{45+7}{9}\right)^{2}-\left(\frac{4\times 9+2}{9}\right)^{2}}$$

Add 45 and 7 to get 52.

$$\frac{\frac{140}{3}}{\left(\frac{52}{9}\right)^{2}-\left(\frac{4\times 9+2}{9}\right)^{2}}$$

Calculate \frac{52}{9} to the power of 2 and get \frac{2704}{81}.

$$\frac{\frac{140}{3}}{\frac{2704}{81}-\left(\frac{4\times 9+2}{9}\right)^{2}}$$

Multiply 4 and 9 to get 36.

$$\frac{\frac{140}{3}}{\frac{2704}{81}-\left(\frac{36+2}{9}\right)^{2}}$$

Add 36 and 2 to get 38.

$$\frac{\frac{140}{3}}{\frac{2704}{81}-\left(\frac{38}{9}\right)^{2}}$$

Calculate \frac{38}{9} to the power of 2 and get \frac{1444}{81}.

$$\frac{\frac{140}{3}}{\frac{2704}{81}-\frac{1444}{81}}$$

Subtract \frac{1444}{81} from \frac{2704}{81} to get \frac{140}{9}.

$$\frac{\frac{140}{3}}{\frac{140}{9}}$$

Divide \frac{140}{3} by \frac{140}{9} by multiplying \frac{140}{3} by the reciprocal of \frac{140}{9}.

$$\frac{140}{3}\times \frac{9}{140}$$

Multiply \frac{140}{3} and \frac{9}{140} to get 3.

$$3$$

Factor

$$3$$