# Bài 19 trang 82 SGK Đại số và Giải tích 12 Nâng cao

##### Hướng dẫn giải

a) $${a^{ - 2\sqrt 2 }}{\left( {{1 \over {{a^{ - \sqrt 2 - 1}}}}} \right)^{\sqrt 2 + 1}} = {a^{ - 2\sqrt 2 }}{\left( {{a^{\sqrt 2 + 1}}} \right)^{\sqrt 2 + 1}} = {a^{ - 2\sqrt 2 }}{a^{3 + 2\sqrt 2 }} = {a^3}$$

b) $${\left( {{{{a^{\sqrt 3 }}} \over {{b^{\sqrt 3 - 1}}}}} \right)^{\sqrt 3 + 1}}{{{a^{ - 1 - \sqrt 3 }}} \over {{b^{ - 2}}}} = {{{a^{3 + \sqrt 3 }}} \over {{b^2}}}.{{{a^{ - 1 - \sqrt 3 }}} \over {{b^{ - 2}}}} = {a^2}$$

c) $${{{a^{2\sqrt 2 }} - {b^{2\sqrt 3 }}} \over {{{\left( {{a^{\sqrt 2 }} - {b^{\sqrt 3 }}} \right)}^2}}} + 1 = {{{a^{2\sqrt 2 }} - {b^{2\sqrt 3 }} + {{\left( {{a^{\sqrt 2 }} - {b^{\sqrt 3 }}} \right)}^2}} \over {{{\left( {{a^{\sqrt 2 }} - {b^{\sqrt 3 }}} \right)}^2}}}$$

$$= {{2{a^{2\sqrt 2 }} - 2{a^{\sqrt 2 }}{b^{\sqrt 3 }}} \over {{{\left( {{a^{\sqrt 2 }} - {b^{\sqrt 3 }}} \right)}^2}}} = {{2{a^{\sqrt 2 }}\left( {{a^{\sqrt 2 }} - {b^{\sqrt 3 }}} \right)} \over {{{\left( {{a^{\sqrt 2 }} - {b^{\sqrt 3 }}} \right)}^2}}} = {{2{a^{\sqrt 2 }}} \over {{a^{\sqrt 2 }} - {b^{\sqrt 3 }}}}$$

d) $$\sqrt {{{\left( {{x^\pi } + {y^\pi }} \right)}^2} - {{\left( {{4^{{1 \over \pi }}}xy} \right)}^\pi }} = \sqrt {{x^{2\pi }} + {y^{2\pi }} - 2{x^\pi }{y^\pi }} = \sqrt {{{\left( {{x^\pi } - {y^\pi }} \right)}^2}} = \left| {{x^\pi } - {y^\pi }} \right|$$.