Bài 103819

Question

Rút gọn biểu thức sau :
$A=\frac{1}{a^2}\sqrt{(a^6+2a^4b^2+3a^2b^4+b^6)^{\frac{2}{3} }}+\left[ {\frac{\left ( b^{\frac{2}{3} }-a^{\frac{2}{3} }\right )^3-2a^2-b^2}{a^2+\left ( b^{\frac{2}{3} }-a^{\frac{2}{3} }\right )^3+2b^2} } \right] $

score 0 · Answer 1

Giải
Ta có : ${a^6} + 3{{\rm{a}}^4}{b^2} + 3{{\rm{a}}^2}{b^4} + {b^6} = {\left( {{a^2} + {b^2}}
\right)^3}$
$\left[ {\frac{{{{\left( {{b^{\frac{2}{3}}} - {a^{\frac{2}{3}}}} \right)}^3} -
2{{\rm{a}}^2} - {b^2}}}{{{a^2} + {{\left( {{b^{\frac{2}{3}}} - {a^{\frac{2}{3}}}} \right)}^3}
+ 2{b^2}}}} \right] = {\left[ {\frac{{{a^2} + {b^2} - 3{b^{\frac{4}{3}}}{a^{\frac{2}{3}}} +
3{{\rm{a}}^{\frac{2}{3}}}{b^{\frac{4}{3}}} - {a^2} + 2{b^2}}}{{{b^2} -
3{{\rm{a}}^{\frac{4}{3}}}{b^{\frac{2}{3}}} + {b^{\frac{2}{3}}}{a^{\frac{4}{3}}} - {a^2} -
2{{\rm{a}}^2} - {b^2}}}} \right]^3}$
$ = \frac{{{b^2} -
{b^{\frac{4}{3}}}{a^{\frac{2}{3}}} + {b^{\frac{2}{3}}}{a^{\frac{4}{3}}}}}{{ - {a^2} -
{b^{\frac{4}{3}}}{a^{\frac{2}{3}}} + {b^{\frac{2}{3}}}{a^{\frac{4}{3}}}}}$
$\Rightarrow $ $A = 1 + \frac{{{b^2}}}{{{a^2}}} + {\left[ {\frac{{{b^{\frac{2}{3}}}\left(
{{b^{\frac{4}{3}}} - {b^{\frac{2}{3}}}{a^{\frac{2}{3}}} + {a^{\frac{4}{3}}}} \right)}}{{ -
{a^{\frac{2}{3}}}\left( {{a^{\frac{4}{3}}} + {b^{\frac{4}{3}}} -
{b^{\frac{2}{3}}}{a^{\frac{2}{3}}}} \right)}}} \right]^3} =1$

Bài 103819

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Bài 103819

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