Solve frac{(2^{2}-2)^{2}}{2^2+2-12}geq8^2-4^2+1+2/9-4^2

$$\frac { ( 2 ^ { 2 } - 2 ) ^ { 2 } } { 2 ^ { 2 + 2 } - 12 } \geq \frac { 8 ^ { 2 } - 4 ^ { 2 + 1 } + 2 } { 9 - 4 ^ { 2 } }$$

Verify

Solution Steps

Calculate 2 to the power of 2 and get 4.

$$\frac{\left(4-2\right)^{2}}{2^{2+2}-12}\geq \frac{8^{2}-4^{2+1}+2}{9-4^{2}}$$

Subtract 2 from 4 to get 2.

$$\frac{2^{2}}{2^{2+2}-12}\geq \frac{8^{2}-4^{2+1}+2}{9-4^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{4}{2^{2+2}-12}\geq \frac{8^{2}-4^{2+1}+2}{9-4^{2}}$$

Add 2 and 2 to get 4.

$$\frac{4}{2^{4}-12}\geq \frac{8^{2}-4^{2+1}+2}{9-4^{2}}$$

Calculate 2 to the power of 4 and get 16.

$$\frac{4}{16-12}\geq \frac{8^{2}-4^{2+1}+2}{9-4^{2}}$$

Subtract 12 from 16 to get 4.

$$\frac{4}{4}\geq \frac{8^{2}-4^{2+1}+2}{9-4^{2}}$$

Divide 4 by 4 to get 1.

$$1\geq \frac{8^{2}-4^{2+1}+2}{9-4^{2}}$$

Calculate 8 to the power of 2 and get 64.

$$1\geq \frac{64-4^{2+1}+2}{9-4^{2}}$$

Add 2 and 1 to get 3.

$$1\geq \frac{64-4^{3}+2}{9-4^{2}}$$

Calculate 4 to the power of 3 and get 64.

$$1\geq \frac{64-64+2}{9-4^{2}}$$

Subtract 64 from 64 to get 0.

$$1\geq \frac{0+2}{9-4^{2}}$$

Add 0 and 2 to get 2.

$$1\geq \frac{2}{9-4^{2}}$$

Calculate 4 to the power of 2 and get 16.

$$1\geq \frac{2}{9-16}$$

Subtract 16 from 9 to get -7.

$$1\geq \frac{2}{-7}$$

Fraction \frac{2}{-7} can be rewritten as -\frac{2}{7} by extracting the negative sign.

$$1\geq -\frac{2}{7}$$

Convert 1 to fraction \frac{7}{7}.

$$\frac{7}{7}\geq -\frac{2}{7}$$

Compare \frac{7}{7} and -\frac{2}{7}.

$$\text{true}$$