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

Bài 19. Đơn giản biểu thức

a) \({a^{ – 2\sqrt 2 }}{\left( {{1 \over {{a^{ – \sqrt 2  – 1}}}}} \right)^{\sqrt 2  + 1}}\);         

b) \({\left( {{{{a^{\sqrt 3 }}} \over {{b^{\sqrt 3  – 1}}}}} \right)^{\sqrt 3  + 1}}{{{a^{ – 1 – \sqrt 3 }}} \over {{b^{ – 2}}}};\)

c) \({{{a^{2\sqrt 2 }} – {b^{2\sqrt 3 }}} \over {{{\left( {{a^{\sqrt 2 }} – {b^{\sqrt 3 }}} \right)}^2}}} + 1;\) 

d) \(\sqrt {{{\left( {{x^\pi } + {y^\pi }} \right)}^2} – {{\left( {{4^{{1 \over \pi }}}xy} \right)}^\pi }} ;\)

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