Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có : \(B=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}\)
Mà \(\frac{1}{2^2}<\frac{1}{1.2};\frac{1}{3^2}<\frac{1}{2.3};...;\frac{1}{8^2}<\frac{1}{7.8}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}<\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}=1-\frac{1}{8}<1\)
Vậy B < 1
\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right)..................\left(1-\frac{1}{20}\right)\)
=\(\frac{1}{2}.\frac{2}{3}.............\frac{19}{20}\)
=\(\frac{1.2.3..............19}{2.3.4..............20}\)
=\(\frac{1}{20}\)
\(A=\frac{\sqrt{x}-5}{\sqrt{x}+5}=\frac{\sqrt{x}+5-10}{\sqrt{x}+5}=1-\frac{10}{\sqrt{x}+5}\)
Vì \(A< \frac{1}{3}=>1-\frac{10}{\sqrt{x}+5}< \frac{1}{3}\)
\(=>1-\frac{1}{3}< \frac{10}{\sqrt{x}+5}=>\frac{2}{3}< \frac{10}{\sqrt{x}+5}\)
\(=>2.\left(\sqrt{x}+5\right)< 30=>2\sqrt{x}+10< 30=>2\sqrt{x}< 20\)
\(=>\sqrt{x}< 10=>\left(\sqrt{x}\right)^2< 10^2=>x< 100\)
Vậy x<100 thì A<1/3
bạn ơi, mình biết làm bài này nhưng cho mình biết làm sao để viết phân số vậy
S=10/2.12+10/12.22+10/22.32+10/32.42+.......+10/2002.2012
S=1/2-1/12+1/12-1/22+1/22-1/32+1/32-1/42+.....+1/2002-1/2012
S=1/2-1/2012
S=????
bạn tự tính nhé
S=10.1/10{1/2-1/12+1/12-1/22+1/22-1/32+...+1/2002-1/2012}
=1/2-1/2012
=1005/2012
Không chép lại đề nhé
Ta có:
P=\(\frac{50-49}{49}+\frac{50-48}{48}+...+\frac{50-2}{2}+\frac{50-1}{1}\)
P=\(\frac{50}{49}-\frac{49}{49}+\frac{50}{48}-\frac{48}{48}+...+\frac{50}{2}-\frac{2}{2}+\frac{50}{1}-\frac{1}{1}\)
P=\(\left(\frac{50}{49}+\frac{50}{48}+...+\frac{50}{2}\right)+\frac{50}{1}-\left(\frac{49}{49}+\frac{48}{48}+...+\frac{2}{2}+\frac{1}{1}\right)\)
P=\(50\cdot\left(\frac{1}{49}+\frac{1}{48}+...+\frac{1}{2}\right)+50-49\) (chỗ này gộp nha)
P=\(50\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{48}+\frac{1}{49}\right)+1\)
P=\(50\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}\right)+\frac{50}{50}\)
P=\(50\cdot\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)\)
=>P=50S
=>\(\frac{S}{P}=\frac{S}{50S}=\frac{1}{50}\)
Vừa nãy mình nói nhầm, Sorry.
Câu 1 :\(P=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).....\left(1-\frac{1}{99}\right)=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{98}{100}=\frac{1}{100}\)
\(S = \frac{1}{3} +\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28} \)
\(S=\frac{1}{3}+\frac{1}{3}.\frac{1}{2}+\frac{1}{5}.\frac{1}{2}+\frac{1}{5}.\frac{1}{3}+\frac{1}{7}.\frac{1}{3}+\frac{1}{7}.\frac{1}{4} \)
\(S=\frac{1}{3}(1+\frac{1}{2})+\frac{1}{5}(\frac{1}{2}+\frac{1}{3})+\frac{1}{7}(\frac{1}{3}+\frac{1}{4})\)
\(S=\frac{1}{3}.\frac{3}{2}+\frac{1}{5}.\frac{5}{6}+\frac{1}{7}.\frac{7}{12}\)
\(S=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}\)
\(S=\frac{6}{12}+\frac{2}{12}+\frac{1}{12}\)
\(S=\frac{9}{12}\)
\(S=\frac{3}{4}\)
S=\(\frac{3}{4}\)