1. Khối 6
  2. Bài tập nâng cao và một số chuyên đề Toán 6 - Tập một
  3. Chương I SỐ TỰ NHIÊN
  4. Chủ đề 3: Luỹ thừa với số mũ tự nhiên. Nhân, chia hai luỹ thừa cùng cơ số
  5. Bài 1

  • Rút gọn các biểu thức sau: a) A=87⋅921185⋅67⋅272A=\dfrac{8^{\tiny 7} \cdot 9^{\tiny 21}}{18^{\tiny 5} \cdot 6^{\tiny 7} \cdot 27^{\tiny 2}}A=185⋅67⋅27287⋅921​
    Trả lời
    b) B=186⋅212⋅43⋅93163⋅69⋅273B=\dfrac{18^{\tiny 6} \cdot 2^{\tiny 12} \cdot 4^{\tiny 3} \cdot 9^{\tiny 3}}{16^{\tiny 3} \cdot 6^{\tiny 9} \cdot 27^{\tiny 3}}B=163⋅69⋅273186⋅212⋅43⋅93​
    c) C=250−249−248−…−22−2C=2^{\tiny 50}-2^{\tiny 49}-2^{\tiny 48}-\ldots-2^{\tiny 2}-2C=250−249−248−…−22−2.

    Đáp án

    Giải thích
    a) Ta có: A=87⋅921185⋅67⋅272=221⋅34225⋅310⋅27⋅37⋅36=221⋅342212⋅323=29⋅319. A=\dfrac{8^{\tiny 7} \cdot 9^{\tiny 21}}{18^{\tiny 5} \cdot 6^{\tiny 7} \cdot 27^{\tiny 2}}=\dfrac{2^{\tiny 21} \cdot 3^{\tiny 42}}{2^{\tiny 5} \cdot 3^{\tiny 10} \cdot 2^{\tiny 7} \cdot 3^{\tiny 7} \cdot 3^{\tiny 6}}=\dfrac{2^{\tiny 21} \cdot 3^{\tiny 42}}{2^{\tiny 12} \cdot 3^{\tiny 23}}=2^{\tiny 9} \cdot 3^{\tiny 19} . A=185⋅67⋅27287⋅921​=25⋅310⋅27⋅37⋅36221⋅342​=212⋅323221⋅342​=29⋅319. b) Ta có: B=186⋅212⋅43⋅93163⋅69⋅273=(2⋅32)6⋅212⋅(22)3⋅(32)3(24)3⋅(2⋅3)9⋅(33)3=26⋅312⋅212⋅26⋅36212⋅29⋅39⋅39=224⋅318221⋅318=23=8. B =\dfrac{18^{\tiny 6} \cdot 2^{\tiny 12} \cdot 4^{\tiny 3} \cdot 9^{\tiny 3}}{16^{\tiny 3} \cdot 6^{\tiny 9} \cdot 27^{\tiny 3}}=\dfrac{\left(2 \cdot 3^{\tiny 2}\right)^{\tiny 6} \cdot 2^{\tiny 12} \cdot\left(2^{\tiny 2}\right)^{\tiny 3} \cdot\left(3^{\tiny 2}\right)^{\tiny 3}}{\left(2^{\tiny 4}\right)^{\tiny 3} \cdot(2 \cdot 3)^{\tiny 9} \cdot\left(3^{\tiny 3}\right)^{\tiny 3}} \\ =\dfrac{2^{\tiny 6} \cdot 3^{\tiny 12} \cdot 2^{\tiny 12} \cdot 2^{\tiny 6} \cdot 3^{\tiny 6}}{2^{\tiny 12} \cdot 2^{\tiny 9} \cdot 3^{\tiny 9} \cdot 3^{\tiny 9}}=\dfrac{2^{\tiny 24} \cdot 3^{\tiny 18}}{2^{\tiny 21} \cdot 3^{\tiny 18}}=2^{\tiny 3}=8 . B=163⋅69⋅273186⋅212⋅43⋅93​=(24)3⋅(2⋅3)9⋅(33)3(2⋅32)6⋅212⋅(22)3⋅(32)3​=212⋅29⋅39⋅3926⋅312⋅212⋅26⋅36​=221⋅318224⋅318​=23=8. c) Ta có: C=250−249−248−…−22−2=250−(249+248+…+22+2)=250−D.C =2^{\tiny 50}-2^{\tiny 49}-2^{\tiny 48}-\ldots-2^{\tiny 2}-2 \\ =2^{\tiny 50}-\left(2^{\tiny 49}+2^{\tiny 48}+\ldots+2^{\tiny 2}+2\right) \\ =2^{\tiny 50}-D . C=250−249−248−…−22−2=250−(249+248+…+22+2)=250−D. Ta có: D=249+248+…+22+2D=2^{\tiny 49}+2^{\tiny 48}+\ldots+2^{\tiny 2}+2D=249+248+…+22+2 suy ra 2D=250+249+248+…+222 D=2^{\tiny 50}+2^{\tiny 49}+2^{\tiny 48}+\ldots+2^{\tiny 2}2D=250+249+248+…+22. Do đó: 2D−D=250−22 D-D=2^{\tiny 50}-22D−D=250−2 nên D=250−2D=2^{\tiny 50}-2D=250−2. Vậy C=250−(250−2)=2C=2^{\tiny 50}-\left(2^{\tiny 50}-2\right)=2C=250−(250−2)=2.
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