$\frac{c^{{\displaystyle \frac{3}{2}}}}{c^{{\displaystyle \frac{1}{2}}}+b^{{\displaystyle \frac{1}{2}}}}—\frac{c{b}^{{\displaystyle \frac{1}{2}}}}{b^{\frac{1}{2}}—c^{{\displaystyle \frac{1}{2}}}}+\frac{2c^2—4cb}{c—b}.$

Приведем первые две дроби к общему знаменателю

$\frac{c^{{\displaystyle \frac{3}{2}}}}{c^{{\displaystyle \frac{1}{2}}}+b^{{\displaystyle \frac{1}{2}}}}—\frac{c{b}^{{\displaystyle \frac{1}{2}}}}{b^{\frac{1}{2}}—c^{{\displaystyle \frac{1}{2}}}}=\frac{c^{{\displaystyle \frac{3}{2}}}\left(b^{\frac{1}{2}}—c^{\frac{1}{2}}\right)}{\left(c^{{\displaystyle \frac{1}{2}}}+b^{{\displaystyle \frac{1}{2}}}\right)\left(b^{\frac{1}{2}}—c^{\frac{1}{2}}\right)}—\frac{c{b}^{{\displaystyle \frac{1}{2}}}\left(c^{\frac{1}{2}}+b^{\frac{1}{2}}\right)}{\left(b^{\frac{1}{2}}—c^{{\displaystyle \frac{1}{2}}}\right)\left(c^{\frac{1}{2}}+b^{\frac{1}{2}}\right)}=\frac{c^{\frac{3}{2}}\left(b^{\frac{1}{2}}—c^{\frac{1}{2}}\right)—c{b}^{{\displaystyle \frac{1}{2}}}\left(c^{\frac{1}{2}}+b^{\frac{1}{2}}\right)}{\left(b^{\frac{1}{2}}—c^{{\displaystyle \frac{1}{2}}}\right)\left(c^{\frac{1}{2}}+b^{\frac{1}{2}}\right)}=\frac{c^{\frac{3}{2}}{b}^{\frac{1}{2}}—c^{\frac{3}{2}}{c}^{\frac{1}{2}}—c{b}^{{\displaystyle \frac{1}{2}}}{c}^{\frac{1}{2}}—c{b}^{\frac{1}{2}}{b}^{\frac{1}{2}}}{\left(b^{\frac{1}{2}}—c^{{\displaystyle \frac{1}{2}}}\right)\left(c^{\frac{1}{2}}+b^{\frac{1}{2}}\right)}.$

Воспользуемся свойством степеней

$a^n·a^m=a^{n+m}$

и формулой сокращенного умножения

$a^2—b^2=\left(a—b\right)\left(a+b\right).$ $\frac{c^{\frac{3}{2}}{b}^{\frac{1}{2}}—c^{\frac{3}{2}}{c}^{\frac{1}{2}}—c{b}^{{\displaystyle \frac{1}{2}}}{c}^{\frac{1}{2}}—c{b}^{\frac{1}{2}}{b}^{\frac{1}{2}}}{\left(b^{\frac{1}{2}}—c^{{\displaystyle \frac{1}{2}}}\right)\left(c^{\frac{1}{2}}+b^{\frac{1}{2}}\right)}=\frac{c^{\frac{3}{2}}{b}^{\frac{1}{2}}—c^{\frac{3}{2}+{\displaystyle \frac{1}{2}}}—c^{1+{\displaystyle \frac{1}{2}}}{b}^{\frac{1}{2}}—c{b}^{\frac{1}{2}+{\displaystyle \frac{1}{2}}}}{b—c}=\frac{c^{\frac{3}{2}}{b}^{\frac{1}{2}}—c^2—c^{{\displaystyle \frac{3}{2}}}{b}^{\frac{1}{2}}—cb}{b—c}=\frac{—c^2—cb}{—\left(c—b\right)}=\frac{—\left(c^2+cb\right)}{—\left(c—b\right)}=\frac{c^2+cb}{c—b}.$ $\frac{c^2+cb}{c—b}+\frac{2c^2—4cb}{c—b}=\frac{c^2+cb+2c^2—4cb}{c—b}=\frac{3c^2—3cb}{c—b}=\frac{3c\left(c—b\right)}{c—b}=3c.$