Bài 8 trang 78 sgk đại số và giải tích 12 nâng cao

\[ = \frac{{\left[ {\sqrt[3]{a} - \sqrt[3]{b}} \right]\left[ {\sqrt[3]{{{a^2}}} + \sqrt[3]{{ab}} + \sqrt[3]{{{b^2}}}} \right]}}{{\left[ {\sqrt[3]{a} - \sqrt[3]{b}} \right]}}\] \[ - \frac{{\left[ {\sqrt[3]{a} + \sqrt[3]{b}} \right]\left[ {\sqrt[3]{{{a^2}}} - \sqrt[3]{{ab}} + \sqrt[3]{{{b^2}}}} \right]}}{{\left[ {\sqrt[3]{a} + \sqrt[3]{b}} \right]}}\]
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  • LG a
  • LG b
  • LG c
  • LG d

Đơn giản biểu thức:

LG a

\[{{\sqrt a - \sqrt b } \over {\root 4 \of a - \root 4 \of b }} - {{\sqrt a + \root 4 \of {ab} } \over {\root 4 \of a + \root 4 \of b }}\]

Lời giải chi tiết:

\[{{\sqrt a - \sqrt b } \over {\root 4 \of a - \root 4 \of b }} - {{\sqrt a + \root 4 \of {ab} } \over {\root 4 \of a + \root 4 \of b }}\]

\[= {{\left[ {\root 4 \of a + \root 4 \of b } \right]\left[ {\root 4 \of a - \root 4 \of b } \right]} \over {\root 4 \of a - \root 4 \of b }} - {{\root 4 \of a \left[ {\root 4 \of a + \root 4 \of b } \right]} \over {\root 4 \of a + \root 4 \of b }}\]

\[ = \root 4 \of a + \root 4 \of b - \root 4 \of a = \root 4 \of b \]

LG b

\[{{a - b} \over {\root 3 \of a - \root 3 \of b }} - {{a + b} \over {\root 3 \of a + \root 3 \of b }}\]

Lời giải chi tiết:

\[{{a - b} \over {\root 3 \of a - \root 3 \of b }} - {{a + b} \over {\root 3 \of a + \root 3 \of b }} \]

\[= {{{{\left[ {\root 3 \of a } \right]}^3} - {{\left[ {\root 3 \of b } \right]}^3}} \over {\root 3 \of a - \root 3 \of b }} - {{{{\left[ {\root 3 \of a } \right]}^3} + {{\left[ {\root 3 \of b } \right]}^3}} \over {\root 3 \of a + \root 3 \of b }}\]

\[ = \frac{{\left[ {\sqrt[3]{a} - \sqrt[3]{b}} \right]\left[ {\sqrt[3]{{{a^2}}} + \sqrt[3]{{ab}} + \sqrt[3]{{{b^2}}}} \right]}}{{\left[ {\sqrt[3]{a} - \sqrt[3]{b}} \right]}}\] \[ - \frac{{\left[ {\sqrt[3]{a} + \sqrt[3]{b}} \right]\left[ {\sqrt[3]{{{a^2}}} - \sqrt[3]{{ab}} + \sqrt[3]{{{b^2}}}} \right]}}{{\left[ {\sqrt[3]{a} + \sqrt[3]{b}} \right]}}\]

\[ = [\root 3 \of {{a^2}} + \root 3 \of {ab} + \root 3 \of {{b^2}}] \] \[ - \left[ {\root 3 \of {{a^2}} - \root 3 \of {ab} + \root 3 \of {{b^2}} } \right] \]

\[= 2\root 3 \of {ab} \]

LG c

\[\left[ {{{a + b} \over {\root 3 \of a + \root 3 \of b }} - \root 3 \of {ab} } \right]:{\left[ {\root 3 \of a - \root 3 \of b } \right]^2};\]

Lời giải chi tiết:

\[\left[ {{{a + b} \over {\root 3 \of a + \root 3 \of b }} - \root 3 \of {ab} } \right]:{\left[ {\root 3 \of a - \root 3 \of b } \right]^2}\]

\[ = [\frac{{\left[ {\sqrt[3]{a} + \sqrt[3]{b}} \right]\left[ {\sqrt[3]{{{a^2}}} - \sqrt[3]{{ab}} + \sqrt[3]{{{b^2}}}} \right]}}{{\left[ {\sqrt[3]{a} + \sqrt[3]{b}} \right]}}- \root 3 \of {ab}]:\] \[:{\left[ {\sqrt[3]{a} - \sqrt[3]{b}} \right]^2}\]

\[= \left[ {\root 3 \of {{a^2}} - \root 3 \of {ab} + \root 3 \of {{b^2}} - \root 3 \of {ab} } \right]:\] \[:{\left[ {\root 3 \of a - \root 3 \of b } \right]^2}\]

\[ = \left[ {\root 3 \of {{a^2}} - 2\root 3 \of {ab} + \root 3 \of {{b^2}} } \right]:{\left[ {\root 3 \of a - \root 3 \of b } \right]^2} \]

\[= {\left[ {\root 3 \of a - \root 3 \of b } \right]^2}:{\left[ {\root 3 \of a - \root 3 \of b } \right]^2} = 1\]

LG d

\[{{a - 1} \over {{a^{{3 \over 4}}} + {a^{{1 \over 2}}}}}.{{\sqrt a + \root 4 \of a } \over {\sqrt a + 1}}.{a^{{1 \over 4}}} + 1.\]

Lời giải chi tiết:

\[{{a - 1} \over {{a^{{3 \over 4}}} + {a^{{1 \over 2}}}}}.{{\sqrt a + \root 4 \of a } \over {\sqrt a + 1}}.{a^{{1 \over 4}}} + 1. \]

\[= {{\left[ {\sqrt a + 1} \right]\left[ {\sqrt a - 1} \right]} \over {\sqrt[4]{{{a^3}}} + \sqrt a }}.{{\root 4 \of a \left[ {\root 4 \of a + 1} \right]} \over {\left[ {\sqrt a + 1} \right]}}.\root 4 \of a + 1\]

\[ = \frac{{\left[ {\sqrt a + 1} \right]\left[ {\sqrt[4]{a} + 1} \right]\left[ {\sqrt[4]{a} - 1} \right]}}{{\sqrt a \left[ {\sqrt[4]{a} + 1} \right]}}.\frac{{\sqrt[4]{a}\left[ {\sqrt[4]{a} + 1} \right]}}{{\sqrt a + 1}}.\sqrt[4]{a} + 1\]

\[ = \frac{{\left[ {\sqrt[4]{a} + 1} \right]\left[ {\sqrt[4]{a} - 1} \right].{{\left[ {\sqrt[4]{a}} \right]}^2}}}{{\sqrt a }} + 1\]

\[ = \sqrt a - 1 + 1 = \sqrt a \].

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