Solve frac{(2^{2}2/3+34/7)^2+[22/3-34/7]^2}{(frac{8}{3})^2+[frac{25}{7}]^2}

$$\frac { ( 2 ^ { 2 } \frac { 2 } { 3 } + 3 \frac { 4 } { 7 } ) ^ { 2 } + [ 2 \frac { 2 } { 3 } - 3 \frac { 4 } { 7 } ] ^ { 2 } } { ( \frac { 8 } { 3 } ) ^ { 2 } + [ \frac { 25 } { 7 } ] ^ { 2 } }$$

Evaluate

$$2$$

Solution Steps

Calculate 2 to the power of 2 and get 4.

$$\frac{\left(4\times \frac{2}{3}+\frac{3\times 7+4}{7}\right)^{2}+\left(\frac{2\times 3+2}{3}-\frac{3\times 7+4}{7}\right)^{2}}{\left(\frac{8}{3}\right)^{2}+\left(\frac{25}{7}\right)^{2}}$$

Multiply 4 and \frac{2}{3} to get \frac{8}{3}.

$$\frac{\left(\frac{8}{3}+\frac{3\times 7+4}{7}\right)^{2}+\left(\frac{2\times 3+2}{3}-\frac{3\times 7+4}{7}\right)^{2}}{\left(\frac{8}{3}\right)^{2}+\left(\frac{25}{7}\right)^{2}}$$

Multiply 3 and 7 to get 21.

$$\frac{\left(\frac{8}{3}+\frac{21+4}{7}\right)^{2}+\left(\frac{2\times 3+2}{3}-\frac{3\times 7+4}{7}\right)^{2}}{\left(\frac{8}{3}\right)^{2}+\left(\frac{25}{7}\right)^{2}}$$

Add 21 and 4 to get 25.

$$\frac{\left(\frac{8}{3}+\frac{25}{7}\right)^{2}+\left(\frac{2\times 3+2}{3}-\frac{3\times 7+4}{7}\right)^{2}}{\left(\frac{8}{3}\right)^{2}+\left(\frac{25}{7}\right)^{2}}$$

Add \frac{8}{3} and \frac{25}{7} to get \frac{131}{21}.

$$\frac{\left(\frac{131}{21}\right)^{2}+\left(\frac{2\times 3+2}{3}-\frac{3\times 7+4}{7}\right)^{2}}{\left(\frac{8}{3}\right)^{2}+\left(\frac{25}{7}\right)^{2}}$$

Calculate \frac{131}{21} to the power of 2 and get \frac{17161}{441}.

$$\frac{\frac{17161}{441}+\left(\frac{2\times 3+2}{3}-\frac{3\times 7+4}{7}\right)^{2}}{\left(\frac{8}{3}\right)^{2}+\left(\frac{25}{7}\right)^{2}}$$

Multiply 2 and 3 to get 6.

$$\frac{\frac{17161}{441}+\left(\frac{6+2}{3}-\frac{3\times 7+4}{7}\right)^{2}}{\left(\frac{8}{3}\right)^{2}+\left(\frac{25}{7}\right)^{2}}$$

Add 6 and 2 to get 8.

$$\frac{\frac{17161}{441}+\left(\frac{8}{3}-\frac{3\times 7+4}{7}\right)^{2}}{\left(\frac{8}{3}\right)^{2}+\left(\frac{25}{7}\right)^{2}}$$

Multiply 3 and 7 to get 21.

$$\frac{\frac{17161}{441}+\left(\frac{8}{3}-\frac{21+4}{7}\right)^{2}}{\left(\frac{8}{3}\right)^{2}+\left(\frac{25}{7}\right)^{2}}$$

Add 21 and 4 to get 25.

$$\frac{\frac{17161}{441}+\left(\frac{8}{3}-\frac{25}{7}\right)^{2}}{\left(\frac{8}{3}\right)^{2}+\left(\frac{25}{7}\right)^{2}}$$

Subtract \frac{25}{7} from \frac{8}{3} to get -\frac{19}{21}.

$$\frac{\frac{17161}{441}+\left(-\frac{19}{21}\right)^{2}}{\left(\frac{8}{3}\right)^{2}+\left(\frac{25}{7}\right)^{2}}$$

Calculate -\frac{19}{21} to the power of 2 and get \frac{361}{441}.

$$\frac{\frac{17161}{441}+\frac{361}{441}}{\left(\frac{8}{3}\right)^{2}+\left(\frac{25}{7}\right)^{2}}$$

Add \frac{17161}{441} and \frac{361}{441} to get \frac{17522}{441}.

$$\frac{\frac{17522}{441}}{\left(\frac{8}{3}\right)^{2}+\left(\frac{25}{7}\right)^{2}}$$

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

$$\frac{\frac{17522}{441}}{\frac{64}{9}+\left(\frac{25}{7}\right)^{2}}$$

Calculate \frac{25}{7} to the power of 2 and get \frac{625}{49}.

$$\frac{\frac{17522}{441}}{\frac{64}{9}+\frac{625}{49}}$$

Add \frac{64}{9} and \frac{625}{49} to get \frac{8761}{441}.

$$\frac{\frac{17522}{441}}{\frac{8761}{441}}$$

Divide \frac{17522}{441} by \frac{8761}{441} by multiplying \frac{17522}{441} by the reciprocal of \frac{8761}{441}.

$$\frac{17522}{441}\times \frac{441}{8761}$$

Multiply \frac{17522}{441} and \frac{441}{8761} to get 2.

$$2$$

Factor

$$2$$