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simplify to get result given by mathematica
- Thread starter xoninhas
- Start date
Expand out the top and bottom, then factor:
\(\displaystyle \frac{1-xy+xy+y^{2}}{1-2xy+x^{2}y^{2}+x^{2}+2xy+y^{2}}\)
\(\displaystyle \frac{1+y^{2}}{1+x^{2}+y^{2}+x^{2}y^{2}}\)
Group and factor the denominator:
\(\displaystyle \frac{1+y^{2}}{(1+y^{2})+x^{2}(1+y^{2})}\)
\(\displaystyle \frac{1+y^{2}}{(1+x^{2})(1+y^{2})}\)
\(\displaystyle \frac{1}{1+x^{2}}\)
\(\displaystyle \frac{1-xy+xy+y^{2}}{1-2xy+x^{2}y^{2}+x^{2}+2xy+y^{2}}\)
\(\displaystyle \frac{1+y^{2}}{1+x^{2}+y^{2}+x^{2}y^{2}}\)
Group and factor the denominator:
\(\displaystyle \frac{1+y^{2}}{(1+y^{2})+x^{2}(1+y^{2})}\)
\(\displaystyle \frac{1+y^{2}}{(1+x^{2})(1+y^{2})}\)
\(\displaystyle \frac{1}{1+x^{2}}\)
Hi !
\(\displaystyle (1-xy)+(x+y)y=1-xy+xy+y^2=1+y^2\)
\(\displaystyle (1-xy)^2+(x+y)^2=(1+(xy)^2-2xy)+(x^2+2xy+y^2)=1+x^2y^2-2xy+2xy+x^2+y^2\)
\(\displaystyle =(1+x^2)+(x^2y^2+y^2)=(1+x^2)+y^2(1+x^2)=(1+x^2)(1+y^2)\)
Therefore :
\(\displaystyle \frac{(1-xy)+(x+y)y}{(1-xy)^2+(x+y)^2}=\frac{1+y^2}{(1+x^2)(1+y^2)}=\boxed{\frac{1}{1+x^2}}\)
\(\displaystyle (1-xy)+(x+y)y=1-xy+xy+y^2=1+y^2\)
\(\displaystyle (1-xy)^2+(x+y)^2=(1+(xy)^2-2xy)+(x^2+2xy+y^2)=1+x^2y^2-2xy+2xy+x^2+y^2\)
\(\displaystyle =(1+x^2)+(x^2y^2+y^2)=(1+x^2)+y^2(1+x^2)=(1+x^2)(1+y^2)\)
Therefore :
\(\displaystyle \frac{(1-xy)+(x+y)y}{(1-xy)^2+(x+y)^2}=\frac{1+y^2}{(1+x^2)(1+y^2)}=\boxed{\frac{1}{1+x^2}}\)
Typo someplace?
Your original expression comes out as...
\(\displaystyle \frac{(1-xy)+(x+y)\cdot y}{(1-xy)^2+(x+y)^2}=\frac{1-xy+xy+y^2}{1-2xy+x^2y^2+x^2+2xy+y^2}=\)
\(\displaystyle \frac{1+y^2}{1+x^2y^2+x^2+y^2}=\frac{1+y^2}{(1+y^2)+x^2(1+y^2)}=\frac{1+y^2}{(1+y^2)(1+x^2)}=\frac{1}{1+x^2}\)
Corrected, thanks to potato
Your original expression comes out as...
\(\displaystyle \frac{(1-xy)+(x+y)\cdot y}{(1-xy)^2+(x+y)^2}=\frac{1-xy+xy+y^2}{1-2xy+x^2y^2+x^2+2xy+y^2}=\)
\(\displaystyle \frac{1+y^2}{1+x^2y^2+x^2+y^2}=\frac{1+y^2}{(1+y^2)+x^2(1+y^2)}=\frac{1+y^2}{(1+y^2)(1+x^2)}=\frac{1}{1+x^2}\)
Corrected, thanks to potato
But denominator is :mrgreen:Loren said:Typo someplace?
Your original expression comes out as...
\(\displaystyle \frac{(1-xy)+(x+y)\cdot y}{(1-xy)^2+(x+y)^2}=\frac{1-xy+xy+y^2}{1-2xy+x^2y^2+x^2+2xy+y^2}=\frac{1+y^2}{1+x^2y^2+x^2+y^2}\)
Numerator is not factorable.
mathematica says:
1/(1+x^2)
xoninhas,
The mathematica solution you provided only works if y = 0. Is there some part of the problem you have not stated?