bài 1 tính nhanh a)18/9.17/37.95/18.-37/34.(-12) b)18/31.15/43+18/31.28/43+18/-31 c) S=1/1.2+1/2.3+1/3.4+….+1/99.100 d ) H=3/1.3+3/3.5+3/5.7+…+3/9

bài 1 tính nhanh a)18/9.17/37.95/18.-37/34.(-12) b)18/31.15/43+18/31.28/43+18/-31 c) S=1/1.2+1/2.3+1/3.4+….+1/99.100 d ) H=3/1.3+3/3.5+3/5.7+…+3/97.99 e) i=91-1/4).(1-1/9).(1-1/10).91-1/100)

0 bình luận về “bài 1 tính nhanh a)18/9.17/37.95/18.-37/34.(-12) b)18/31.15/43+18/31.28/43+18/-31 c) S=1/1.2+1/2.3+1/3.4+….+1/99.100 d ) H=3/1.3+3/3.5+3/5.7+…+3/9”

  1. Đáp án:

     $1)
    a) \dfrac{190}{3}\\
    b)
    0\\
    c)
    S=\dfrac{99}{100}\\
    d)
    H=\dfrac{49}{33}\\
    e)
    i=\dfrac{297}{500}$

    Giải thích các bước giải:

     $1)
    a) \dfrac{18}{9}.\dfrac{17}{37}.\dfrac{95}{18}.\dfrac{-37}{34}.(-12)\\
    =2.\dfrac{17}{37}.\dfrac{-37}{34}.\dfrac{95}{18}.(-12)\\
    =2.\dfrac{-17}{34}.\dfrac{95}{3}.(-2)\\
    =2.\dfrac{-1}{2}.\dfrac{-190}{3}\\
    =\dfrac{190}{3}\\
    b)
    \dfrac{18}{31}.\dfrac{15}{43}+\dfrac{18}{31}.\dfrac{28}{43}+\dfrac{18}{-31}\\
    =\dfrac{18}{31}.\left ( \dfrac{15}{43}+\dfrac{28}{43}-1 \right )\\
    =\dfrac{18}{31}.\left ( \dfrac{43}{43}-1 \right )\\
    =\dfrac{18}{31}.0=0\\
    c)
    S=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+…+\dfrac{1}{99.100}\\
    =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+…+\dfrac{1}{99}-\dfrac{1}{100}\\
    =1-\dfrac{1}{100}\\
    =\dfrac{100}{100}-\dfrac{1}{100}\\
    =\dfrac{99}{100}\\
    d)
    H=\dfrac{3}{1.3}+\dfrac{3}{3.5}+\dfrac{3}{5.7}+…+\dfrac{3}{97.99}\\
    =3.\left ( \dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+…+\dfrac{1}{97.99} \right )\\
    =\dfrac{3}{2}.\left (\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+…+\dfrac{2}{97.99}  \right )\\
    =\dfrac{3}{2}.\left ( \dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+…+\dfrac{1}{97}-\dfrac{1}{99} \right )\\
    =\dfrac{3}{2}.\left ( 1-\dfrac{1}{99} \right )\\
    =\dfrac{3}{2}.\left ( \dfrac{99}{99}-\dfrac{1}{99} \right )\\
    =\dfrac{3}{2}.\dfrac{98}{99}\\
    =\dfrac{49}{33}\\
    e)
    i=\left (1-\dfrac{1}{4}  \right ).\left ( 1-\dfrac{1}{9} \right ).\left ( 1-\dfrac{1}{10} \right ).\left ( 1-\dfrac{1}{100} \right )\\
    =\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{9}{10}.\dfrac{99}{100}\\
    =\dfrac{3.8.9.99}{4.9.10.100}\\
    =\dfrac{3.2.99}{10.100}\\
    =\dfrac{297}{500}$

    Bình luận

Viết một bình luận