Giải thích các bước giải:

a,

ĐKXĐ:  \(a \ne  \pm b \ne 0\)

Ta có:

\(\begin{array}{l}
P = \dfrac{{{a^2}}}{{ab + {b^2}}} + \dfrac{{{b^2}}}{{ab - {a^2}}} - \dfrac{{{a^2} + {b^2}}}{{ab}}\\
 = \dfrac{{{a^2}}}{{b\left( {a + b} \right)}} + \dfrac{{{b^2}}}{{a\left( {b - a} \right)}} - \dfrac{{{a^2} + {b^2}}}{{ab}}\\
 = \dfrac{{{a^2}}}{{b\left( {a + b} \right)}} - \dfrac{{{b^2}}}{{a\left( {a - b} \right)}} - \dfrac{{{a^2} + {b^2}}}{{ab}}\\
 = \dfrac{{{a^3}\left( {a - b} \right) - {b^3}\left( {a + b} \right) - \left( {{a^2} + {b^2}} \right)\left( {a - b} \right)\left( {a + b} \right)}}{{ab\left( {a - b} \right)\left( {a + b} \right)}}\\
 = \dfrac{{{a^4} - {a^3}b - a{b^3} - {b^4} - \left( {{a^2} + {b^2}} \right)\left( {{a^2} - {b^2}} \right)}}{{ab\left( {a - b} \right)\left( {a + b} \right)}}\\
 = \dfrac{{{a^4} - {a^3}b - a{b^3} - {b^4} - \left( {{a^4} - {b^4}} \right)}}{{ab\left( {a - b} \right)\left( {a + b} \right)}}\\
 = \dfrac{{ - {a^3}b - a{b^3}}}{{ab\left( {a - b} \right)\left( {a + b} \right)}}\\
 = \dfrac{{ - ab\left( {{a^2} + {b^2}} \right)}}{{ab\left( {a - b} \right)\left( {a + b} \right)}}\\
 = \dfrac{{ - \left( {{a^2} + {b^2}} \right)}}{{{a^2} - {b^2}}}\\
b,\\
P = 0 \Leftrightarrow \dfrac{{ - \left( {{a^2} + {b^2}} \right)}}{{{a^2} - {b^2}}} = 0 \Leftrightarrow {a^2} + {b^2} = 0 \Rightarrow a = b = 0
\end{array}\)

Từ ĐKXĐ suy ra không có giá trị nào của a,b thỏa mãn P=0

\(\begin{array}{l}
c,\\
3{a^2} + 3{b^2} = 10ab\\
 \Leftrightarrow 3{a^2} - 10ab + 3{b^2} = 0\\
 \Leftrightarrow \left( {3{a^2} - 9ab} \right) - \left( {ab - 3{b^2}} \right) = 0\\
 \Leftrightarrow 3a\left( {a - 3b} \right) - b\left( {a - 3b} \right) = 0\\
 \Leftrightarrow \left( {a - 3b} \right)\left( {3a - b} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
a = 3b\\
3a = b
\end{array} \right.\\
TH1:\,\,\,a = 3b\\
P = \dfrac{{ - \left( {{a^2} + {b^2}} \right)}}{{{a^2} - {b^2}}} = \dfrac{{ - \left( {9{b^2} + {b^2}} \right)}}{{9{b^2} - {b^2}}} = \dfrac{{ - 10{b^2}}}{{8{b^2}}} =  - \dfrac{5}{4}\\
TH2:\,\,\,b = 3a\\
P = \dfrac{{ - \left( {{a^2} + {b^2}} \right)}}{{{a^2} - {b^2}}} = \dfrac{{ - \left( {{a^2} + 9{a^2}} \right)}}{{{a^2} - 9{a^2}}} = \dfrac{{ - 10{a^2}}}{{ - 8{a^2}}} = \dfrac{5}{4}
\end{array}\)