Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
\dfrac{{\sqrt a  + \sqrt b }}{{\sqrt a  - \sqrt b }} + \dfrac{{\sqrt a  - \sqrt b }}{{\sqrt a  + \sqrt b }}\\
 = \dfrac{{{{\left( {\sqrt a  + \sqrt b } \right)}^2} + {{\left( {\sqrt a  - \sqrt b } \right)}^2}}}{{\left( {\sqrt a  - \sqrt b } \right)\left( {\sqrt a  + \sqrt b } \right)}}\\
 = \dfrac{{\left( {a + 2\sqrt {ab}  + b} \right) + \left( {a - 2\sqrt {ab}  + b} \right)}}{{a - b}}\\
 = \dfrac{{2.\left( {a + b} \right)}}{{a - b}}\\
b,\\
\dfrac{{a - b}}{{\sqrt a  - \sqrt b }} - \dfrac{{\sqrt {{a^3}}  - \sqrt {{b^3}} }}{{a - b}}\\
 = \dfrac{{\left( {\sqrt a  - \sqrt b } \right)\left( {\sqrt a  + \sqrt b } \right)}}{{\sqrt a  - \sqrt b }} - \dfrac{{\left( {\sqrt a  - \sqrt b } \right)\left( {a + \sqrt {ab}  + b} \right)}}{{\left( {\sqrt a  - \sqrt b } \right)\left( {\sqrt a  + \sqrt b } \right)}}\\
 = \sqrt a  + \sqrt b  - \dfrac{{a + \sqrt {ab}  + b}}{{\sqrt a  + \sqrt b }}\\
 = \dfrac{{{{\left( {\sqrt a  + \sqrt b } \right)}^2} - \left( {a + \sqrt {ab}  + b} \right)}}{{\sqrt a  + \sqrt b }}\\
 = \dfrac{{\left( {a + 2\sqrt {ab}  + b} \right) - \left( {a + \sqrt {ab}  + b} \right)}}{{\sqrt a  + \sqrt b }}\\
 = \dfrac{{\sqrt {ab} }}{{\sqrt a  + \sqrt b }}
\end{array}\)

Giải thích các bước giải:

 a)$\frac{√a+√b}{√a-√b}$+ $\frac{√a-√b}{√a+√b}$(a,b≥0,a$\neq$ b)

=$\frac{(√a+√b)²}{a-b}$+ $\frac{(√a-√b)²}{a-b}$

=$\frac{a+2√ab+b+a-2√ab+b}{a-b}$=$\frac{2a+2b}{a-b}$

=$\frac{2(a+b)}{a-b}$=2(a,b≥0,a$\neq$ b)

b)$\frac{a-b}{√a-√b}$-$\frac{√a³-√b³}{a-b}$(a,b≥0,a$\neq$ b)

=$\frac{(a-b)(√a+√b)}{a-b}$-$\frac{√a³-√b³}{a-b}$

=$\frac{√a³+a√b-b√a-√b³-√a³+√b³}{a-b}$

=$\frac{a√b-b√a}{a-b}$=$\frac{√ab}{√a+√b}$(a,b≥0,a$\neq$ b)