So sanh A va B
A = 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 +1/8 + 1/9 +1/10
B = 23/10
Tính giúp em với ạ
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
DT
Đoàn Thảo Nhi
6 tháng 2 2023
Đúng(0)
Những câu hỏi liên quan
17 tháng 5 2022
a: \(=\dfrac{-3}{7}+\dfrac{15}{26}-\dfrac{2}{13}+\dfrac{3}{7}=\dfrac{15}{26}-\dfrac{4}{26}=\dfrac{11}{26}\)
b: \(=\dfrac{6}{7}+\dfrac{2}{9}-\dfrac{10}{7}-5=\dfrac{-4}{7}-5+\dfrac{2}{9}=-\dfrac{337}{63}\)
c: \(=-\dfrac{11}{23}\left(\dfrac{6}{7}+\dfrac{8}{7}\right)-\dfrac{1}{23}=\dfrac{-22}{23}-\dfrac{1}{23}=-1\)
PT
7 tháng 5 2017
Giải:
a) \(\left(-\dfrac{5}{28}+1,75+\dfrac{8}{35}\right):\left(\dfrac{-39}{20}\right)\)
\(=\left(-\dfrac{5}{28}+\dfrac{7}{4}+\dfrac{8}{35}\right):\left(\dfrac{-39}{20}\right)\)
\(=\left(\dfrac{11}{7}+\dfrac{8}{35}\right):\left(-\dfrac{39}{20}\right)\)
\(=\dfrac{9}{5}:\left(-\dfrac{39}{20}\right)\)
\(=\dfrac{9.\left(-20\right)}{5.39}\)
\(=\dfrac{3.\left(-4\right)}{1.13}\)
\(=\dfrac{-12}{13}\)
b) \(6\dfrac{5}{12}:2\dfrac{3}{4}+11\dfrac{1}{4}.\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{77}{22}:\dfrac{11}{4}+\dfrac{42}{4}.\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{77}{22}.\dfrac{4}{11}+\dfrac{42}{4}.\left(\dfrac{5}{15}+\dfrac{3}{15}\right)\)
\(=\dfrac{7}{3}+\dfrac{42}{4}.\dfrac{8}{15}\)
\(=\dfrac{7}{3}+\dfrac{14.2}{1.3}\)
\(=\dfrac{7}{3}+\dfrac{28}{3}\)
\(=\dfrac{35}{3}\)
c) \(70,5-528:\dfrac{15}{2}\)
\(=70,5-528.\dfrac{2}{15}\)
\(=70,5-\dfrac{1056}{15}\)
\(=70,5-70,4\)
\(=0,1\)
2 tháng 5 2017
a) \(\dfrac{2}{3}x-\dfrac{3}{2}x=\dfrac{5}{12}\)
\(x\left(\dfrac{2}{3}-\dfrac{3}{2}\right)=\dfrac{5}{12}\)
\(x\cdot\left(-\dfrac{5}{6}\right)=\dfrac{5}{12}\)
\(x=\dfrac{5}{12}:\left(-\dfrac{5}{6}\right)\)
\(x=-\dfrac{1}{2}\)
Vậy \(x=-\dfrac{1}{2}\).
b) \(\dfrac{2}{5}+\dfrac{3}{5}\cdot\left(3x-3\cdot7\right)=-\dfrac{53}{10}\)
\(\dfrac{3}{5}\left(3x-3\cdot7\right)=-\dfrac{53}{10}-\dfrac{2}{5}\)
\(\dfrac{3}{5}\left(3x-3\cdot7\right)=-\dfrac{57}{10}\)
\(3x-3\cdot7=-\dfrac{57}{10}:\dfrac{3}{5}\)
\(3x-3\cdot7=-\dfrac{19}{2}\)
\(3x-21=-\dfrac{19}{2}\)
\(3x=-\dfrac{19}{2}+21\)
\(3x=\dfrac{23}{2}\)
\(x=\)\(\dfrac{23}{2}:3\)
\(x=\dfrac{23}{6}\)
Vậy \(x=\dfrac{23}{6}\).
c) \(\dfrac{7}{9}:\left(2+\dfrac{3}{4x}\right)+\dfrac{5}{3}=\dfrac{23}{27}\)
\(\dfrac{7}{9}:\left(2+\dfrac{3}{4x}\right)=\dfrac{23}{27}-\dfrac{5}{3}\)
\(\dfrac{7}{9}:\left(2+\dfrac{3}{4x}\right)=-\dfrac{22}{27}\)
\(2+\dfrac{3}{4x}=\dfrac{7}{9}:-\dfrac{22}{27}\)
\(2+\dfrac{3}{4x}=-\dfrac{21}{22}\)
\(\dfrac{3}{4x}=-\dfrac{21}{22}-2\)
\(\dfrac{3}{4x}=-\dfrac{65}{22}\)
\(4x=\dfrac{3\cdot22}{-65}\)
\(4x=-\dfrac{66}{65}\)
\(x=-\dfrac{66}{65}:4\)
\(x=-\dfrac{33}{130}\)
Vậy \(x=-\dfrac{33}{130}\).
d) \(-\dfrac{2}{3}x+\dfrac{1}{5}=\dfrac{3}{10}\)
\(-\dfrac{2}{3}x=\dfrac{3}{10}-\dfrac{1}{5}\)
\(-\dfrac{2}{3}x=\dfrac{1}{10}\)
\(x=\dfrac{1}{10}:-\dfrac{2}{3}\)
\(x=-\dfrac{3}{20}\)
Vậy \(x=-\dfrac{3}{20}\).
e) \(\left|x\right|-\dfrac{3}{4}=\dfrac{5}{3}\)
\(\left|x\right|=\dfrac{5}{3}+\dfrac{3}{4}\)
\(\left|x\right|=\dfrac{29}{12}\)
\(x=\dfrac{29}{12}\) hoặc \(=-\dfrac{29}{12}\)
Vậy \(x\in\left\{\dfrac{29}{12};-\dfrac{29}{12}\right\}\).
VD
2
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
23 tháng 8 2016
\(A< \frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}=\frac{5}{5}=1=B\)
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
23 tháng 8 2016
a/
\(\frac{2001}{2004}=\frac{2004-3}{2004}=1-\frac{3}{2004}=1-\frac{1}{668}.\)
\(\frac{39}{40}=\frac{40-1}{40}=1-\frac{1}{40}\)
Ta có \(40< 668\Rightarrow\frac{1}{40}>\frac{1}{668}\Rightarrow1-\frac{1}{40}< 1-\frac{1}{668}\Rightarrow\frac{39}{40}< \frac{2001}{2004}\)
b/
\(A< \frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}=1=B\)
H
2 tháng 1 2018
\(A=\frac{1+5+5^2+...+5^9}{1+5+5^2+...+5^8}=\frac{1+5\left(1 +5+5^2+...+5^8\right)}{1+5+5^2+...+5^8}=5+\frac{1}{1+5+5^2+...+5^8} \)
\(B=\frac{1+3+3^2+....+3^9}{1+3+3^2+....+3^8}=\frac{1+3\left(1+3+3^2+....+3^8\right)}{1+3+3^2+....+3^8}=3+\frac{1}{1+3+3^2+....+3^8}\)
\(=5+\frac{1}{1+3+3^2+....+3^8}-2\)
Có: \(\frac{1}{1+5+5^2+...+5^8}>0\) và \(\frac{1}{1+3+3^2+....+3^8}-2< 0\)
\(\Rightarrow A>B\)
M
0
