- LG a
- LG b
- LG c
- LG d
Đơn giản biểu thức
LG a
\[{a^{ - 2\sqrt 2 }}{\left[ {{1 \over {{a^{ - \sqrt 2 - 1}}}}} \right]^{\sqrt 2 + 1}}\];
Lời giải chi tiết:
\[{a^{ - 2\sqrt 2 }}{\left[ {{1 \over {{a^{ - \sqrt 2 - 1}}}}} \right]^{\sqrt 2 + 1}} \]
\[ = {a^{ - 2\sqrt 2 }}.{\left[ {{{\left[ {{a^{ - \sqrt 2 - 1}}} \right]}^{ - 1}}} \right]^{\sqrt 2 + 1}}\]
\[= {a^{ - 2\sqrt 2 }}{\left[ {{a^{\sqrt 2 + 1}}} \right]^{\sqrt 2 + 1}} \]
\[ = {a^{ - 2\sqrt 2 }}.{a^{\left[ {\sqrt 2 + 1} \right]\left[ {\sqrt 2 + 1} \right]}}\]
\[= {a^{ - 2\sqrt 2 }}{a^{3 + 2\sqrt 2 }} = {a^{ - 2\sqrt 2 + 3 + 2\sqrt 2 }}\]
\[= {a^3}\]
LG 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:
\[{\left[ {{{{a^{\sqrt 3 }}} \over {{b^{\sqrt 3 - 1}}}}} \right]^{\sqrt 3 + 1}}{{{a^{ - 1 - \sqrt 3 }}} \over {{b^{ - 2}}}} \]
\[ = \frac{{{{\left[ {{a^{\sqrt 3 }}} \right]}^{\sqrt 3 + 1}}}}{{{{\left[ {{b^{\sqrt 3 - 1}}} \right]}^{\sqrt 3 + 1}}}}.\frac{{{a^{ - 1 - \sqrt 3 }}}}{{{b^{ - 2}}}} \]
\[= \frac{{{a^{\sqrt 3 .\left[ {\sqrt 3 + 1} \right]}}}}{{{b^{\left[ {\sqrt 3 - 1} \right]\left[ {\sqrt 3 + 1} \right]}}}}.\frac{{{a^{ - 1 - \sqrt 3 }}}}{{{b^{ - 2}}}}\]
\[= {{{a^{3 + \sqrt 3 }}} \over {{b^2}}}.{{{a^{ - 1 - \sqrt 3 }}} \over {{b^{ - 2}}}} \]
\[= \frac{{{a^{3 + \sqrt 3 }}.{a^{ - 1 - \sqrt 3 }}}}{{{b^2}.{b^{ - 2}}}} = \frac{{{a^{3 + \sqrt 3 - 1 - \sqrt 3 }}}}{{{b^{2 - 2}}}} = \frac{{{a^2}}}{{{b^0}}}= {a^2}\]
LG c
\[{{{a^{2\sqrt 2 }} - {b^{2\sqrt 3 }}} \over {{{\left[ {{a^{\sqrt 2 }} - {b^{\sqrt 3 }}} \right]}^2}}} + 1;\]
Lời giải chi tiết:
\[{{{a^{2\sqrt 2 }} - {b^{2\sqrt 3 }}} \over {{{\left[ {{a^{\sqrt 2 }} - {b^{\sqrt 3 }}} \right]}^2}}} + 1 = {{{a^{2\sqrt 2 }} - {b^{2\sqrt 3 }} + {{\left[ {{a^{\sqrt 2 }} - {b^{\sqrt 3 }}} \right]}^2}} \over {{{\left[ {{a^{\sqrt 2 }} - {b^{\sqrt 3 }}} \right]}^2}}}\]
\[ = \frac{{{a^{2\sqrt 2 }} - {b^{2\sqrt 3 }} + {a^{2\sqrt 2 }} - 2{a^{\sqrt 2 }}.{b^{\sqrt 3 }} + {b^{2\sqrt 3 }}}}{{{{\left[ {{a^{\sqrt 2 }} - {b^{\sqrt 3 }}} \right]}^2}}}\]
\[ = {{2{a^{2\sqrt 2 }} - 2{a^{\sqrt 2 }}{b^{\sqrt 3 }}} \over {{{\left[ {{a^{\sqrt 2 }} - {b^{\sqrt 3 }}} \right]}^2}}} \]
\[= {{2{a^{\sqrt 2 }}\left[ {{a^{\sqrt 2 }} - {b^{\sqrt 3 }}} \right]} \over {{{\left[ {{a^{\sqrt 2 }} - {b^{\sqrt 3 }}} \right]}^2}}}\]
\[= {{2{a^{\sqrt 2 }}} \over {{a^{\sqrt 2 }} - {b^{\sqrt 3 }}}}\]
LG d
\[\sqrt {{{\left[ {{x^\pi } + {y^\pi }} \right]}^2} - {{\left[ {{4^{{1 \over \pi }}}xy} \right]}^\pi }} ;\]
Lời giải chi tiết:
\[\sqrt {{{\left[ {{x^\pi } + {y^\pi }} \right]}^2} - {{\left[ {{4^{{1 \over \pi }}}xy} \right]}^\pi }} \]
\[\begin{array}{l}
= \sqrt {{{\left[ {{x^\pi }} \right]}^2} + 2{x^\pi }{y^\pi } + {{\left[ {{y^\pi }} \right]}^2} - {{\left[ {{4^{\frac{1}{\pi }}}} \right]}^\pi }{x^\pi }{y^\pi }} \\
= \sqrt {{x^{2\pi }} + 2{x^\pi }{y^\pi } + {y^{2\pi }} - 4{x^\pi }{y^\pi }}
\end{array}\]
\[= \sqrt {{x^{2\pi }} + {y^{2\pi }} - 2{x^\pi }{y^\pi }} \]
\[= \sqrt {{{\left[ {{x^\pi } - {y^\pi }} \right]}^2}} \]
\[= \left| {{x^\pi } - {y^\pi }} \right|\].