S = \(\dfrac{1}{5^2}+\dfrac{1}{7^2}+\dfrac{1}{9^2}+...+\dfrac{1}{103^2}\) CMR:S ⋮\(\dfrac{5}{32}\)
Giúp em với ạ . Mai em thi rồi ạ
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\dfrac{2}{3}+\dfrac{1}{5}.\dfrac{10}{7}=\dfrac{2}{3}+\dfrac{10}{35}=\dfrac{70}{105}+\dfrac{30}{105}=\dfrac{100}{105}=\dfrac{50}{21}\)
a) Ta có: \(\dfrac{2}{3}+\dfrac{1}{5}\cdot\dfrac{10}{7}\)
\(=\dfrac{2}{3}+\dfrac{2}{7}\)
\(=\dfrac{14}{21}+\dfrac{6}{21}\)
\(=\dfrac{20}{21}\)
a) \(\dfrac{49}{81}=\dfrac{7^x}{9^x}\)(sửa đề)
\(\Leftrightarrow\left(\dfrac{7}{9}\right)^2=\left(\dfrac{7}{9}\right)^x\)\(\Rightarrow x=2\)
b) \(\dfrac{-64}{343}=\left(-\dfrac{4^x}{7^x}\right)\)(sửa đề)
\(\Leftrightarrow\left(-\dfrac{4}{7}\right)^3=\left(-\dfrac{4}{7}\right)^x\) \(\Rightarrow x=3\)
c) \(\dfrac{9}{144}=\dfrac{3^x}{12^x}\)(sửa đề)
\(\Leftrightarrow\left(\dfrac{3}{12}\right)^2=\left(\dfrac{3}{12}\right)^x\Rightarrow x=2\)
d) \(-\dfrac{1}{32}=\left(-\dfrac{1^x}{2^x}\right)\)(sửa đề)
\(\Leftrightarrow\left(-\dfrac{1}{2}\right)^5=\left(-\dfrac{1}{2}\right)^x\Rightarrow x=5\)
Mong bạn xem lại đề bài.
\(49,=\dfrac{38}{5}-\left(\dfrac{19}{7}+\dfrac{28}{5}\right)\)
\(=\dfrac{38}{5}-\dfrac{19}{7}-\dfrac{28}{5}\)
\(=\left(\dfrac{38}{5}-\dfrac{28}{5}\right)-\dfrac{19}{7}\)
\(=2-\dfrac{19}{7}=-\dfrac{5}{7}\)
\(50,=\dfrac{25}{81}.\dfrac{15}{22}=\dfrac{125}{594}\)
mik xinloi ạ câu 50 mik viết sai đề nên mong bạn giải lại giúp ạ <3
Ta có \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};...;\dfrac{1}{20^2}< \dfrac{1}{19.20}\)
Cộng vế với vế ta được
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{20^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{19}-\dfrac{1}{20}\)
\(\Rightarrow T< 2-\dfrac{1}{20}=\dfrac{39}{20}\)
mà 39/20 < 8/7 => T < 8/7
\(-2\dfrac{1}{4}.\)\(\left(3\dfrac{5}{12}-1\dfrac{2}{9}\right)\)
=\(\dfrac{-9}{4}\).\(\left(\dfrac{41}{12}-\dfrac{11}{9}\right)\)
=\(\dfrac{-9}{4}.\dfrac{41}{12}-\dfrac{-9}{4}.\dfrac{11}{9}\)
=\(\dfrac{-123}{16}-\dfrac{-11}{4}\)
=\(\dfrac{-123}{16}-\dfrac{-44}{16}\)
=\(\dfrac{-79}{16}\)
\(\left(-25\%+0,75+\dfrac{7}{12}\right)\div\left(-2\dfrac{1}{8}\right)\)
=\(\left(\dfrac{-1}{4}+\dfrac{3}{4}+\dfrac{7}{12}\right)\div\left(\dfrac{-17}{8}\right)\)
=\(\left(\dfrac{-3}{12}+\dfrac{9}{12}+\dfrac{7}{12}\right).\dfrac{-8}{17}\)
=\(\dfrac{13}{12}.\dfrac{-8}{17}=\dfrac{-26}{51}\)
khôngn đăng lại
c#ẹp.c#ẹp