1/6<1/5^2+1/6^2+1/7^2+...+1/100^2<1/4
ta có:
(+)1/5^2+1/6^2+1/7^2+...+1/100^2<1/4.5+1/5.6+...+1/99.100
=1/4-1/5+1/5-...+1/99-1/100
=1/4-1/100<1/4
=>1/5^2+1/6^2+1/7^2+...+1/100^2<1/4
(+)1/5^2+1/6^2+1/7^2+...+1/100^2>1/5.6+...+1/99.100
=1/5-1/6+1/6-...+1/99-1/100
=1/5-1/100>1/6
=>1/5^2+1/6^2+1/7^2+...+1/100^2
Chứng minh rằng
\(\dfrac{1}{6}< \dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4}\)
Chứng minh rằng:
\(\dfrac{1}{7^2}-\dfrac{1}{7^4}+...+\dfrac{1}{7^{4n-2}}-\dfrac{1}{7^{4n}}+...+\dfrac{1}{7^{98}}+\dfrac{1}{7^{100}}< \dfrac{1}{50}\)
1,\(\dfrac{3}{16}\)- ( x - \(\dfrac{5}{4}\) ) - ( \(\dfrac{3}{4}\) + \(\dfrac{-7}{8}\) - 1 ) = \(2\dfrac{1}{2}\)
2,\(\dfrac{1}{2}\) . ( \(\dfrac{1}{6}\) - \(\dfrac{9}{10}\) ) = \(\dfrac{1}{5}\) - x + ( \(\dfrac{1}{15}\) - \(\dfrac{-1}{5}\) )
Giúp mik nhanh với ạ
a) \(\dfrac{\left(x+\dfrac{3}{4}\right)\cdot\dfrac{7}{2}-\dfrac{1}{6}}{-\left(\dfrac{4}{5}+\dfrac{1}{3}\right)\cdot\dfrac{1}{2}+1}=2\dfrac{33}{52}\)
b)\(\dfrac{\left(5-\dfrac{2}{7}\right)\cdot\dfrac{7}{9}\cdot\dfrac{3}{5}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=5\dfrac{5}{21}\)
a)\(\dfrac{\left(x+\dfrac{3}{4}\right)\cdot\dfrac{7}{2}-\dfrac{1}{6}}{-\left(\dfrac{4}{5}+\dfrac{1}{3}\right)\cdot\dfrac{1}{2}+1}=2\dfrac{33}{52}\)
b)\(\dfrac{\left(5-\dfrac{2}{7}\right)\cdot\dfrac{7}{9}:\dfrac{3}{5}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=5\dfrac{5}{21}\)
Bài 7: Tìm X
Cho \(A=1+\dfrac{3}{2^3}+\dfrac{4}{2^4}+\dfrac{5}{2^5}+...+\dfrac{100}{2^{100}}\). Chứng minh A < 2.
Chứng minh rằng : \(B=\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+\dfrac{4}{3^4}+...+\dfrac{100}{3^{100}}< \dfrac{3}{4}\)
\(\dfrac{4}{9}.19\dfrac{1}{3}+\dfrac{-4}{9}.39\dfrac{1}{3}+\dfrac{-3^2}{5}:\left(-0,2\right)\)
\(\dfrac{\left(-5^3\right)^2}{27}.\dfrac{3}{23}.\dfrac{6^2}{-5^6}\)
\(1\dfrac{4}{23}+\dfrac{5}{21}-\dfrac{4}{23}+0,5+\dfrac{16}{21}\)
\(\dfrac{3}{7}.19\dfrac{1}{3}-\dfrac{3}{7}:33\dfrac{1}{3}\)
\(\dfrac{3}{4}-\left(x+\dfrac{1}{2}\right)=\dfrac{4}{5}\)
x:(-2,14)=(-3,12):1,2
Bài 1: Thực hiện phép tính:
\(A=\left(-\dfrac{1}{125}\right)^{11}:\left(\dfrac{1}{5}\right)^{32}\)
\(B=1+\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2+....+\left(\dfrac{1}{3}\right)^{2018}\)
\(C=\dfrac{16^3\cdot3^{10}+120\cdot6^9}{4^6\cdot3^{12}+6^{11}}\)
\(D=\left(\dfrac{0.4-\dfrac{2}{9}+\dfrac{2}{11}}{1.4-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-0.25+\dfrac{1}{5}}{1\dfrac{1}{6}-0.875+0.7}\right):\dfrac{2017}{2018}\)
\(E=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\)
\(G=\dfrac{\left(\dfrac{2}{5}\right)^7\cdot5^7+\left(2\dfrac{1}{4}\right)^3:\left(\dfrac{3}{16}\right)^3}{512+2^7\cdot5^2}:\dfrac{\left(\dfrac{1}{2}\right)^0}{\left(-1\right)^{2017}}\)
Mn ơi giúp e với ........ Em đang cần gấp giúp e với nha!!
Thank you mn nhiều nhiều.....
