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