Solve 20042003^2+1^2/20042002^2+20042004^2

$$\frac { 20042003 ^ { 2 } + 1 ^ { 2 } } { 20042002 ^ { 2 } + 20042004 ^ { 2 } }$$

Evaluate

$$\frac{1}{2}=0.5$$

Solution Steps

Calculate 20042003 to the power of 2 and get 401681884252009.

$$\frac{401681884252009+1^{2}}{20042002^{2}[phone]^{2}}$$

Calculate 1 to the power of 2 and get 1.

$$\frac{401681884252009+1}{20042002^{2}[phone]^{2}}$$

Add 401681884252009 and 1 to get 401681884252010.

$$\frac{401681884252010}{20042002^{2}[phone]^{2}}$$

Calculate 20042002 to the power of 2 and get 401681844168004.

$$\frac{401681884252010}{401681844168004+20042004^{2}}$$

Calculate 20042004 to the power of 2 and get 401681924336016.

$$\frac{401681884252010}{401681844168004+401681924336016}$$

Add 401681844168004 and 401681924336016 to get 803363768504020.

$$\frac{401681884252010}{803363768504020}$$

Reduce the fraction \frac{401681884252010}{803363768504020} to lowest terms by extracting and canceling out 401681884252010.

$$\frac{1}{2}$$

Factor

$$\frac{1}{2} = 0.5$$