A = \(10\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)\)
A : B = 10
Đúng 0
Bình luận (0)
Các câu hỏi tương tự
Tìm thương A:B, biết:
\(A=\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+...+\frac{1}{9}\)
\(B=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\)
Tìm A:B, biết:
A=\(\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+\frac{6}{4}+\frac{5}{5}+\frac{4}{6}+\frac{3}{7}+\frac{2}{8}+\frac{1}{9}\)
B=\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{9}+\frac{1}{10}\)
Giúp tớ làm bài này:
Tìm thương A chia B biết :
A =\(\frac{9}{1}\)+\(\frac{8}{2}+\frac{7}{3}+...+\frac{1}{9}\)
B =\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\)
Tìm x biết
a) (8-5x).(x+2)+4.(x-2).(x+1)+2.(x-2).(x+2)=0
b)\(\left(-\frac{2}{5}+x\right):\frac{7}{9}+\left(-\frac{3}{5}+\frac{5}{6}\right):\frac{7}{9}=0\)
c)\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=1\frac{2003}{2004}\)
Tìm x biết
a) x+2x+3x+4x+...+100x=-213
b)\(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)
c)3(x-2)+2(x-1)=10
d)\(\frac{x+1}{3}=\frac{x-2}{4}\)
e)\(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+\frac{x-11}{12}\)
f)\(\frac{x+32}{11}+\frac{x+23}{12}=\frac{x+38}{13}+\frac{x+27}{14}\)
P=\(\frac{\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+...+\frac{1}{9}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}}\)
Tính P
tính :
a) \(\frac{4}{9}:\frac{-1}{7}+6\frac{5}{9}:\frac{-1}{7}\)
b) \(\left(2\frac{1}{3}+3\frac{1}{2}\right):\left(-4\frac{1}{6}+3\frac{1}{7}\right)+7\frac{1}{2}\)
c) \(\frac{\left(13\frac{1}{4}-2\frac{5}{7}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{7}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
Tính hợp lí: \(A=\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}:\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\)
tính
a, \(\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}\)
b , \(\left(\frac{0,4-\frac{8}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{1\frac{1}{6}-0,875+0,7}\right):\frac{2012}{2013}\)
c, A
= \(1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+\frac{1}{4}.\left(1+2+3+...+\frac{1}{20}.\left(1+2+3+....+20\right)\right).155\)