\(\dfrac{1}{\dfrac{3}{4}+1}=\dfrac{1}{\dfrac{7}{4}}=\dfrac{4}{7}\)
Chúc bạn học tốt!!!
\(\dfrac{1}{\dfrac{3}{4}+1}=\dfrac{1}{\dfrac{7}{4}}=\dfrac{4}{7}\)
Chúc bạn học tốt!!!
1.Thuc hien phap tinh a) \((\)\(2\dfrac{3}{5}\)-\(3\dfrac{5}{9}\)\()\):\((\)\(3\dfrac{10}{21}\)-\(1\dfrac{3}{7}\)\()\) b) \(5\dfrac{1}{2}\)-\(14\dfrac{3}{7}\):\(\dfrac{9}{13}\)-\(3\dfrac{4}{7}\):\(\dfrac{9}{13}\)
1.Tinh gia tri cua bieu thuc
\(\dfrac{1}{2}+\dfrac{1}{3}\).x-\(\dfrac{1}{6}\).x voi x=\(\dfrac{-3}{5}\)
1 tinh
a,\(5\dfrac{4}{23}.27\dfrac{3}{47}+4\dfrac{3}{47}.\left(-5\dfrac{4}{23}\right)\)
b,4.\(\left(\dfrac{-1}{2}\right)^3+\dfrac{3}{2}\)
c,\(\left(\dfrac{1999}{2011}-\dfrac{2011}{1999}\right)-\left(\dfrac{-12}{1999}-\dfrac{12}{2011}\right)\)
d,\(\left(\dfrac{-5}{11}+\dfrac{7}{22}-\dfrac{-4}{33}-\dfrac{5}{44}\right):\left(\dfrac{381}{22}-39\dfrac{7}{22}\right)\)
Thu gọn:
A = \(\dfrac{2^4.3^3+2^3.3^4}{2^5.3^3-2^4.3^2}\) D = \(\dfrac{\left(\dfrac{1}{2}-\dfrac{1}{3}\right).\left(\dfrac{2}{3}-\dfrac{3}{4}\right)}{\left(\dfrac{1}{2}+\dfrac{1}{3}\right).\left(\dfrac{2}{3}+\dfrac{3}{4}\right)}\)
B = \(\dfrac{2^3-3^4-2^4.3^3}{2^5.3^4-2^6.3^3}\) E = \(\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{3}{4}\right).\left(\dfrac{2}{3}-\dfrac{3}{4}+\dfrac{5}{6}\right)\)
C = \(\dfrac{\dfrac{1}{2}-\dfrac{1}{2}:\dfrac{3}{4}-\dfrac{3}{4}}{\dfrac{2}{3}-\dfrac{2}{3}:\dfrac{5}{6}-\dfrac{5}{6}}\)
Giúp mình với ! 10k nha
tinh:
a. \(\dfrac{3}{4}-1\dfrac{1}{2}+0,5:\dfrac{5}{12}\)
b. \(\left(-2\right)^2-1\dfrac{5}{27}.\left(-\dfrac{3}{2}\right)^3\)
c. \(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\)
BT1: CMR:
a) \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}< 1\)
b) \(\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+\dfrac{1}{64}+\dfrac{1}{100}+\dfrac{1}{144}+\dfrac{1}{196}< \dfrac{1}{2}\)
c) \(\dfrac{1}{3}+\dfrac{1}{30}+\dfrac{1}{32}+\dfrac{1}{35}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{50}< \dfrac{1}{2}\)
d) \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}< \dfrac{1}{3}\)
e) \(\dfrac{1}{3}< \dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}< \dfrac{3}{16}\)
f) \(\dfrac{1}{41}+\dfrac{1}{42}+\dfrac{1}{43}+...+\dfrac{1}{79}+\dfrac{1}{80}>\dfrac{7}{12}\)
BT2: Tính tổng
a) A=\(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}\)
b) E=\(1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+\dfrac{1}{4}\left(1+2+3+4\right)+...+\dfrac{1}{200}\left(1+2+3+...+200\right)\)
BT3: Cho S=\(\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}+\dfrac{3}{13}+\dfrac{3}{14}\)
CMR: 1 < S < 2
tinh: \(\dfrac{3}{4}-1\dfrac{1}{2}+0,5:\dfrac{5}{12}\)
1) \(B=\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+\dfrac{1}{5^3}+...+\dfrac{1}{500^3}< \dfrac{1}{4}\)
2) \(D=\dfrac{4}{3}+\dfrac{10}{9}+\dfrac{28}{27}+...+\dfrac{3^{98}+1}{3^{98}}< 100\)
Tính hợp lí A = \(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}}{\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}\right)-\dfrac{1}{2}.\dfrac{1}{3}.\dfrac{1}{4}}\)