Bài 1:
A=\(\dfrac{\dfrac{5}{12}+\dfrac{3}{4}-1}{3-\dfrac{5}{6}+\dfrac{2}{3}}+\dfrac{\dfrac{16}{5}+\dfrac{16}{7}-\dfrac{16}{9}}{\dfrac{17}{5}+\dfrac{17}{7}-\dfrac{17}{9}}\)
A=\(\dfrac{\dfrac{1}{6}}{\dfrac{17}{6}}+\dfrac{16\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}\right)}{17\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}\right)}\)
A=\(\dfrac{1.\dfrac{1}{6}}{17.\dfrac{1}{6}}+\dfrac{16}{17}\)
A=\(\dfrac{1}{17}+\dfrac{16}{17}=\dfrac{17}{17}=1\)
Bài 2 mk chưa có câu trả lời, sorry nha!
S = \(\dfrac{1}{5}+\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{41}+\dfrac{1}{42}\)
S = \(\dfrac{5932}{12915}\)
S = 0,4593108788... \(\approx\)0,5
Vì \(0,5\) = \(0,5\)
=> \(\dfrac{1}{5}+\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{41}+\dfrac{1}{42}\) = \(\dfrac{1}{2}\)
Có j sai thì nói vs mk nha! đúng thì tick giúp mk nha!
Bài 2 :
S = \(\dfrac{1}{5}+\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{41}+\dfrac{1}{42}\) với \(\dfrac{1}{2}\)
Ta thấy : \(\dfrac{1}{9}+\dfrac{1}{10}< \dfrac{1}{8}+\dfrac{1}{8}=\dfrac{1}{4}\)
Và : \(\dfrac{1}{41}+\dfrac{1}{42}< \dfrac{1}{40}+\dfrac{1}{40}=\dfrac{1}{20}\)
Suy ra : S < \(\dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}=\dfrac{4+5+1}{20}=\dfrac{1}{2}\)