Toán Lớp 8: so sánh A vs B:
a) A = (a + b)(a^2 + b^2)(a^4 + b^4)…..(a^32 + b^32)
B = a^64 – b^64. với a = b + 1
b) A = 2005 × 7( 8^2003 + 8^2002 + …+ 8^2 + 8 + 1 ) + 2005
B = 8^2004 × 2005
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Toán Lớp 8: so sánh A vs B:
a) A = (a + b)(a^2 + b^2)(a^4 + b^4)…..(a^32 + b^32)
B = a^64 – b^64. với a = b + 1
b) A = 2005 × 7( 8^2003 + 8^2002 + …+ 8^2 + 8 + 1 ) + 2005
B = 8^2004 × 2005
Giúp mik với
Comments ( 1 )
a)A = \left( {a + b} \right)\left( {{a^2} + {b^2}} \right)\left( {{a^4} + {b^4}} \right)…\left( {{a^{32}} + {b^{32}}} \right)\\
= \dfrac{1}{{a – b}}.\left( {a – b} \right).\left( {a + b} \right)\left( {{a^2} + {b^2}} \right)\left( {{a^4} + {b^4}} \right)…\left( {{a^{32}} + {b^{32}}} \right)\\
= \dfrac{1}{{a – b}}.\left( {{a^2} – {b^2}} \right)\left( {{a^2} + {b^2}} \right)\left( {{a^4} + {b^4}} \right)…\left( {{a^{32}} + {b^{32}}} \right)\\
= \dfrac{1}{{a – b}}.\left( {{a^4} – {b^4}} \right)\left( {{a^4} + {b^4}} \right)…\left( {{a^{32}} + {b^{32}}} \right)\\
= \dfrac{1}{{a – b}}.\left( {{a^{32}} – {b^{32}}} \right)\left( {{a^{32}} + {b^{32}}} \right)\\
= \dfrac{1}{{a – b}}.\left( {{a^{64}} – {b^{64}}} \right)\\
= {a^{64}} – {b^{64}}\left( {do:a = b + 1 \Leftrightarrow a – b = 1} \right)\\
Vay\,A = B\\
b)A = 2005.7.\left( {{8^{2003}} + {8^{2002}} + … + {8^2} + 8 + 1} \right) + 2005\\
a = \left( {{8^{2003}} + {8^{2002}} + … + {8^2} + 8 + 1} \right)\\
\Leftrightarrow 8a = {8^{2004}} + {8^{2003}} + … + {8^3} + {8^2} + 8\\
\Leftrightarrow 8a – a = {8^{2004}} – 1\\
\Leftrightarrow a = \dfrac{{{8^{2004}} – 1}}{7}\\
\Leftrightarrow A = 2005.7.\dfrac{{{8^{2004}} – 1}}{7} + 2005\\
= 2005.\left( {{8^{2004}} – 1} \right) + 2005\\
= {2005.8^{2004}}\\
\Leftrightarrow A = B
\end{array}$