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HK
1
16 tháng 8 2022
\(A=\dfrac{15\left(1+2\cdot4+64\right)}{35+240+2240}\)
\(=\dfrac{15\cdot73}{2515}=\dfrac{15\cdot73}{5\cdot503}=\dfrac{3\cdot73}{503}=\dfrac{219}{503}>\dfrac{3}{8}\)
20 tháng 4 2019
\(A=\frac{\text{1.5.6 + 2.10.12 + 3.15.18 + 4.20.24 + 5.25.30}}{\text{1.3.5 + 2.6.10 + 3.9.15 + 4.12.20 + 5.15.25 }}\)
\(=\frac{1.5.6+2.\left(1.5.6\right)+3.\left(1.5.6\right)+4.\left(1.5.6\right)+5.\left(1.5.6\right)}{1.3.5+2.\left(1.3.5\right)+3.\left(1.3.5\right)+4.\left(1.3.5\right)+5.\left(1.3.5\right)}\)
\(=\frac{30.\left(1+2+3+4+5\right)}{15.\left(1+2+3+4+5\right)}\)
\(=\frac{30}{15}=2\)
Vậy A=2.
4 tháng 5 2017
\(=\frac{1.5.6+\left(1.5.6\right).2+\left(1.5.6\right).3+\left(1.5.6\right).4+\left(1.5.6\right).5}{1.3.5+\left(1.3.5\right).2+\left(1.3.5\right).3+\left(1.3.5\right).4+\left(1.3.5\right).5}\)
\(=\frac{\left(1.5.6\right).\left(1+2+3+4+5\right)}{\left(1.3.5\right).\left(1+2+3+4+5\right)}=\frac{1.5.6}{1.3.5}=\frac{1.1.2}{1.1.1}=2\)
4 tháng 5 2017
A= \(\frac{1.5.3.2+2.10.2.6+2.15.9.2+4.20.12.2+5.25.15.2}{1.3.5+2.6.10+3.9.15+4.12.20+5.15.25}\)
A= \(\frac{2+2+2\cdot2+2+2}{0+0+3+0+0}\)
A= \(\frac{12}{3}\)
A= 4
Đầu tiên bạn tách ra, rút gọn rồi cộng lại,tính nha!
18 tháng 3 2018
Ta có :
\(S=\frac{3}{2}+\frac{4}{3}+\frac{5}{4}+\frac{6}{5}+\frac{7}{6}+\frac{8}{7}+\frac{9}{8}+\frac{10}{9}+\frac{11}{10}+\frac{12}{11}\)
\(S=\frac{2+1}{2}+\frac{3+1}{3}+\frac{4+1}{4}+...+\frac{11+1}{11}\)
\(S=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{3}\right)+\left(1+\frac{1}{4}\right)+...+\left(1+\frac{1}{11}\right)\)
\(S=\left(1+1+1+...+1\right)+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{11}\right)\)
\(S=10+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{11}\right)>10\)
\(\Rightarrow\)\(S>10\)
Vậy \(S>10\)
Chúc bạn học tốt ~
HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023
a) Ta có: \(\frac{2}{{ - 5}} = \frac{{ - 16}}{{40}}\) và \(\frac{{ - 3}}{8} = \frac{{ - 15}}{{40}}\)
Do \(\frac{{ - 16}}{{40}} < \frac{{ - 15}}{{40}}\,\, \Rightarrow \,\frac{2}{{ - 5}} < \frac{{ - 3}}{8}\).
b) Ta có: \( - 0,85 = \frac{{ - 85}}{{100}} = \frac{{ - 17}}{{20}}\). Vậy \( - 0,85\)=\(\frac{{ - 17}}{{20}}\).
c) Ta có: \(\frac{{37}}{{ - 25}} = \frac{{ - 296}}{{200}}\)
Do \(\frac{{ - 137}}{{200}} > \frac{{ - 296}}{{200}}\) nên \(\frac{{ - 137}}{{200}}\) > \(\frac{{37}}{{ - 25}}\) .
d) Ta có: \( - 1\frac{3}{{10}}=\frac{-13}{10}\) ;
\(-\left( {\frac{{ - 13}}{{ - 10}}} \right) = \frac{{-13}}{{10}}\).
Vậy \(- 1\frac{3}{{10}} =-(\frac{{-13}}{{-10}})\,\).
11 tháng 10 2017
ta có A= \(\frac{8^{18}+1}{8^{19} +1}\)=> 8A=\(\frac{8^{19}+8}{8^{19}+1}\)= \(\frac{\left(8^{19}+1\right)+7}{8^{19}+1}\)=\(\frac{8^{19}+1}{8^{19} +1}\)+\(\frac{7}{8^{19}+1}\) =1+\(\frac{7}{8^{19}+1}\) =\(\frac{7}{8^{19}+1}\)
B= \(\frac{8^{23}+1}{8^{24}+1}\)=> 8B=\(\frac{8^{24}+8}{8^{24}+1}\)= \(\frac{\left(8^{24}+1\right)+7}{8^{24}+1}\)=\(\frac{8^{24}+1}{8^{24}+1}\)+\(\frac{7}{8^{24}+1}\) =1+\(\frac{7}{8^{24} +1}\)=\(\frac{7}{8^{24}+1}\)
vì \(8^{19}\)<\(8^{24}\)=> \(8^{19}\)+1 >\(8^{24}\)+1 => \(\frac{7}{8^{19}+1}\)<\(\frac{7}{8^{24}+1}\)=> A<B
11 tháng 10 2017
a) ta có \(8A=\frac{8^{19}+8}{8^{19}+1}=1+\frac{7}{8^{19}+1}\\ 8B=\frac{8^{24}+8}{8^{24}+1}=1+\frac{7}{8^{24}+1}\)
Vì \(8^{24}+1>8^{19}+1\\\frac{7}{8^{24}+1}< \frac{7}{8^{19}+1} \)
vậy 8A>8B nên A>B
\(\frac{1.3.5+2.6.10+4.12.20}{1.5.7+2.10.14+4.20.28}\)
\(=\frac{3.5+2.3.2.5.2+4.3.4.5.4}{5.7+2.5.2.2.7+4.4.5.7.4}\)
\(=\frac{3.5.\left(1+2.2.2+4.4.4\right)}{5.7.\left(1+2.2.2+4.4.4\right)}\)
\(=\frac{3}{7}>\frac{3}{8}\)