Google
Câu hỏi: Đơn giản biểu thức
Lời giải chi tiết:
\({a^{ - 2\sqrt 2 }}{\left( {{1 \over {{a^{ - \sqrt 2 - 1}}}}} \right)^{\sqrt 2 + 1}} \)
\(= {a^{ - 2\sqrt 2 }}.{\left[ {{{\left( {{a^{ - \sqrt 2 - 1}}} \right)}^{ - 1}}} \right]^{\sqrt 2 + 1}}\)
\(= {a^{ - 2\sqrt 2 }}{\left( {{a^{\sqrt 2 + 1}}} \right)^{\sqrt 2 + 1}} \)
\(= {a^{ - 2\sqrt 2 }}.{a^{\left( {\sqrt 2 + 1} \right)\left({\sqrt 2 + 1} \right)}}\)
\(= {a^{ - 2\sqrt 2 }}{a^{3 + 2\sqrt 2 }} = {a^{ - 2\sqrt 2 + 3 + 2\sqrt 2 }}\)
\(= {a^3}\)
Lời giải chi tiết:
\({\left( {{{{a^{\sqrt 3 }}} \over {{b^{\sqrt 3 - 1}}}}} \right)^{\sqrt 3 + 1}}{{{a^{ - 1 - \sqrt 3 }}} \over {{b^{ - 2}}}} \)
\(= \frac{{{{\left( {{a^{\sqrt 3 }}} \right)}^{\sqrt 3 + 1}}}}{{{{\left({{b^{\sqrt 3 - 1}}} \right)}^{\sqrt 3 + 1}}}}.\frac{{{a^{ - 1 - \sqrt 3 }}}}{{{b^{ - 2}}}} \)
\(= \frac{{{a^{\sqrt 3 .\left( {\sqrt 3 + 1} \right)}}}}{{{b^{\left({\sqrt 3 - 1} \right)\left({\sqrt 3 + 1} \right)}}}}.\frac{{{a^{ - 1 - \sqrt 3 }}}}{{{b^{ - 2}}}}\)
\(= {{{a^{3 + \sqrt 3 }}} \over {{b^2}}}.{{{a^{ - 1 - \sqrt 3 }}} \over {{b^{ - 2}}}} \)
\(= \frac{{{a^{3 + \sqrt 3 }}.{a^{ - 1 - \sqrt 3 }}}}{{{b^2}.{b^{ - 2}}}} = \frac{{{a^{3 + \sqrt 3 - 1 - \sqrt 3 }}}}{{{b^{2 - 2}}}} = \frac{{{a^2}}}{{{b^0}}}= {a^2}\)
Lời giải chi tiết:
\({{{a^{2\sqrt 2 }} - {b^{2\sqrt 3 }}} \over {{{\left( {{a^{\sqrt 2 }} - {b^{\sqrt 3 }}} \right)}^2}}} + 1 = {{{a^{2\sqrt 2 }} - {b^{2\sqrt 3 }} + {{\left({{a^{\sqrt 2 }} - {b^{\sqrt 3 }}} \right)}^2}} \over {{{\left({{a^{\sqrt 2 }} - {b^{\sqrt 3 }}} \right)}^2}}}\)
\(= \frac{{{a^{2\sqrt 2 }} - {b^{2\sqrt 3 }} + {a^{2\sqrt 2 }} - 2{a^{\sqrt 2 }}.{b^{\sqrt 3 }} + {b^{2\sqrt 3 }}}}{{{{\left( {{a^{\sqrt 2 }} - {b^{\sqrt 3 }}} \right)}^2}}}\)
\(= {{2{a^{2\sqrt 2 }} - 2{a^{\sqrt 2 }}{b^{\sqrt 3 }}} \over {{{\left( {{a^{\sqrt 2 }} - {b^{\sqrt 3 }}} \right)}^2}}} \)
\(= {{2{a^{\sqrt 2 }}\left( {{a^{\sqrt 2 }} - {b^{\sqrt 3 }}} \right)} \over {{{\left({{a^{\sqrt 2 }} - {b^{\sqrt 3 }}} \right)}^2}}}\)
\(= {{2{a^{\sqrt 2 }}} \over {{a^{\sqrt 2 }} - {b^{\sqrt 3 }}}}\)
Lời giải chi tiết:
\(\sqrt {{{\left( {{x^\pi } + {y^\pi }} \right)}^2} - {{\left({{4^{{1 \over \pi }}}xy} \right)}^\pi }} \)
\(\begin{array}{l}
= \sqrt {{{\left({{x^\pi }} \right)}^2} + 2{x^\pi }{y^\pi } + {{\left({{y^\pi }} \right)}^2} - {{\left({{4^{\frac{1}{\pi }}}} \right)}^\pi }{x^\pi }{y^\pi }} \\
= \sqrt {{x^{2\pi }} + 2{x^\pi }{y^\pi } + {y^{2\pi }} - 4{x^\pi }{y^\pi }}
\end{array}\)
\(= \sqrt {{x^{2\pi }} + {y^{2\pi }} - 2{x^\pi }{y^\pi }} \)
\(= \sqrt {{{\left( {{x^\pi } - {y^\pi }} \right)}^2}} \)
\(= \left| {{x^\pi } - {y^\pi }} \right|\).
Câu a
\({a^{ - 2\sqrt 2 }}{\left( {{1 \over {{a^{ - \sqrt 2 - 1}}}}} \right)^{\sqrt 2 + 1}}\);Lời giải chi tiết:
\({a^{ - 2\sqrt 2 }}{\left( {{1 \over {{a^{ - \sqrt 2 - 1}}}}} \right)^{\sqrt 2 + 1}} \)
\(= {a^{ - 2\sqrt 2 }}.{\left[ {{{\left( {{a^{ - \sqrt 2 - 1}}} \right)}^{ - 1}}} \right]^{\sqrt 2 + 1}}\)
\(= {a^{ - 2\sqrt 2 }}{\left( {{a^{\sqrt 2 + 1}}} \right)^{\sqrt 2 + 1}} \)
\(= {a^{ - 2\sqrt 2 }}.{a^{\left( {\sqrt 2 + 1} \right)\left({\sqrt 2 + 1} \right)}}\)
\(= {a^{ - 2\sqrt 2 }}{a^{3 + 2\sqrt 2 }} = {a^{ - 2\sqrt 2 + 3 + 2\sqrt 2 }}\)
\(= {a^3}\)
Câu b
\({\left( {{{{a^{\sqrt 3 }}} \over {{b^{\sqrt 3 - 1}}}}} \right)^{\sqrt 3 + 1}}{{{a^{ - 1 - \sqrt 3 }}} \over {{b^{ - 2}}}};\)Lời giải chi tiết:
\({\left( {{{{a^{\sqrt 3 }}} \over {{b^{\sqrt 3 - 1}}}}} \right)^{\sqrt 3 + 1}}{{{a^{ - 1 - \sqrt 3 }}} \over {{b^{ - 2}}}} \)
\(= \frac{{{{\left( {{a^{\sqrt 3 }}} \right)}^{\sqrt 3 + 1}}}}{{{{\left({{b^{\sqrt 3 - 1}}} \right)}^{\sqrt 3 + 1}}}}.\frac{{{a^{ - 1 - \sqrt 3 }}}}{{{b^{ - 2}}}} \)
\(= \frac{{{a^{\sqrt 3 .\left( {\sqrt 3 + 1} \right)}}}}{{{b^{\left({\sqrt 3 - 1} \right)\left({\sqrt 3 + 1} \right)}}}}.\frac{{{a^{ - 1 - \sqrt 3 }}}}{{{b^{ - 2}}}}\)
\(= {{{a^{3 + \sqrt 3 }}} \over {{b^2}}}.{{{a^{ - 1 - \sqrt 3 }}} \over {{b^{ - 2}}}} \)
\(= \frac{{{a^{3 + \sqrt 3 }}.{a^{ - 1 - \sqrt 3 }}}}{{{b^2}.{b^{ - 2}}}} = \frac{{{a^{3 + \sqrt 3 - 1 - \sqrt 3 }}}}{{{b^{2 - 2}}}} = \frac{{{a^2}}}{{{b^0}}}= {a^2}\)
Câu c
\({{{a^{2\sqrt 2 }} - {b^{2\sqrt 3 }}} \over {{{\left( {{a^{\sqrt 2 }} - {b^{\sqrt 3 }}} \right)}^2}}} + 1;\)Lời giải chi tiết:
\({{{a^{2\sqrt 2 }} - {b^{2\sqrt 3 }}} \over {{{\left( {{a^{\sqrt 2 }} - {b^{\sqrt 3 }}} \right)}^2}}} + 1 = {{{a^{2\sqrt 2 }} - {b^{2\sqrt 3 }} + {{\left({{a^{\sqrt 2 }} - {b^{\sqrt 3 }}} \right)}^2}} \over {{{\left({{a^{\sqrt 2 }} - {b^{\sqrt 3 }}} \right)}^2}}}\)
\(= \frac{{{a^{2\sqrt 2 }} - {b^{2\sqrt 3 }} + {a^{2\sqrt 2 }} - 2{a^{\sqrt 2 }}.{b^{\sqrt 3 }} + {b^{2\sqrt 3 }}}}{{{{\left( {{a^{\sqrt 2 }} - {b^{\sqrt 3 }}} \right)}^2}}}\)
\(= {{2{a^{2\sqrt 2 }} - 2{a^{\sqrt 2 }}{b^{\sqrt 3 }}} \over {{{\left( {{a^{\sqrt 2 }} - {b^{\sqrt 3 }}} \right)}^2}}} \)
\(= {{2{a^{\sqrt 2 }}\left( {{a^{\sqrt 2 }} - {b^{\sqrt 3 }}} \right)} \over {{{\left({{a^{\sqrt 2 }} - {b^{\sqrt 3 }}} \right)}^2}}}\)
\(= {{2{a^{\sqrt 2 }}} \over {{a^{\sqrt 2 }} - {b^{\sqrt 3 }}}}\)
Câu d
\(\sqrt {{{\left( {{x^\pi } + {y^\pi }} \right)}^2} - {{\left({{4^{{1 \over \pi }}}xy} \right)}^\pi }} ;\)Lời giải chi tiết:
\(\sqrt {{{\left( {{x^\pi } + {y^\pi }} \right)}^2} - {{\left({{4^{{1 \over \pi }}}xy} \right)}^\pi }} \)
\(\begin{array}{l}
= \sqrt {{{\left({{x^\pi }} \right)}^2} + 2{x^\pi }{y^\pi } + {{\left({{y^\pi }} \right)}^2} - {{\left({{4^{\frac{1}{\pi }}}} \right)}^\pi }{x^\pi }{y^\pi }} \\
= \sqrt {{x^{2\pi }} + 2{x^\pi }{y^\pi } + {y^{2\pi }} - 4{x^\pi }{y^\pi }}
\end{array}\)
\(= \sqrt {{x^{2\pi }} + {y^{2\pi }} - 2{x^\pi }{y^\pi }} \)
\(= \sqrt {{{\left( {{x^\pi } - {y^\pi }} \right)}^2}} \)
\(= \left| {{x^\pi } - {y^\pi }} \right|\).
Rất tiếc, câu hỏi này chưa có lời giải chi tiết. Bạn ơi, đăng nhập và giải chi tiết giúp zix.vn nhé!!!