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\frac{\left(4-9\right)^{2}-\left(-3\right)^{2}}{|7^{2}-54|+3}>\frac{4^{2}-7-\left(11-2\right)}{\left(11-1\right)^{2}+5}
Calculate 3 to the power of 2 and get 9.
\frac{\left(-5\right)^{2}-\left(-3\right)^{2}}{|7^{2}-54|+3}>\frac{4^{2}-7-\left(11-2\right)}{\left(11-1\right)^{2}+5}
Subtract 9 from 4 to get -5.
\frac{25-\left(-3\right)^{2}}{|7^{2}-54|+3}>\frac{4^{2}-7-\left(11-2\right)}{\left(11-1\right)^{2}+5}
Calculate -5 to the power of 2 and get 25.
\frac{25-9}{|7^{2}-54|+3}>\frac{4^{2}-7-\left(11-2\right)}{\left(11-1\right)^{2}+5}
Calculate -3 to the power of 2 and get 9.
\frac{16}{|7^{2}-54|+3}>\frac{4^{2}-7-\left(11-2\right)}{\left(11-1\right)^{2}+5}
Subtract 9 from 25 to get 16.
\frac{16}{|49-54|+3}>\frac{4^{2}-7-\left(11-2\right)}{\left(11-1\right)^{2}+5}
Calculate 7 to the power of 2 and get 49.
\frac{16}{|-5|+3}>\frac{4^{2}-7-\left(11-2\right)}{\left(11-1\right)^{2}+5}
Subtract 54 from 49 to get -5.
\frac{16}{5+3}>\frac{4^{2}-7-\left(11-2\right)}{\left(11-1\right)^{2}+5}
The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of -5 is 5.
\frac{16}{8}>\frac{4^{2}-7-\left(11-2\right)}{\left(11-1\right)^{2}+5}
Add 5 and 3 to get 8.
2>\frac{4^{2}-7-\left(11-2\right)}{\left(11-1\right)^{2}+5}
Divide 16 by 8 to get 2.
2>\frac{16-7-\left(11-2\right)}{\left(11-1\right)^{2}+5}
Calculate 4 to the power of 2 and get 16.
2>\frac{9-\left(11-2\right)}{\left(11-1\right)^{2}+5}
Subtract 7 from 16 to get 9.
2>\frac{9-9}{\left(11-1\right)^{2}+5}
Subtract 2 from 11 to get 9.
2>\frac{0}{\left(11-1\right)^{2}+5}
Subtract 9 from 9 to get 0.
2>\frac{0}{10^{2}+5}
Subtract 1 from 11 to get 10.
2>\frac{0}{100+5}
Calculate 10 to the power of 2 and get 100.
2>\frac{0}{105}
Add 100 and 5 to get 105.
2>0
Zero divided by any non-zero number gives zero.
\text{true}
Compare 2 and 0.