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\(\Leftrightarrow\frac{1}{2}x-\frac{3}{8}-\frac{2}{5}x=\frac{17}{4}\)
\(\Leftrightarrow\frac{1}{2}x-\frac{2}{5}x=\frac{17}{4}+\frac{3}{8}\)(Bạn tự quy đồng chỗ này)
\(\Leftrightarrow\frac{1}{10}x=\frac{37}{8}\)
\(\Leftrightarrow x=\frac{185}{4}\)
\(P=\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+\frac{4}{5^5}+...+\frac{11}{5^{12}}\)
\(\Rightarrow\)\(5P=\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+\frac{4}{5^4}+...+\frac{11}{5^{11}}\)
\(\Rightarrow\)\(4P=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+\frac{1}{5^4}+...+\frac{1}{5^{11}}-\frac{1}{5^{12}}\)
\(\Rightarrow\)\(20P=1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{10}}-\frac{1}{5^{11}}\)
\(\Rightarrow\)\(16P=1-\frac{1}{5^{11}}+\frac{1}{5^{12}}-\frac{1}{5^{11}}\)\(< 1\)
\(\Rightarrow\)\(P< \frac{1}{16}\)
P/s: nguyên tác: https://olm.vn/thanhvien/nhatphuonghocgiot
Ta có
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{11^2}>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{11.12}\)
Mà
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{11.12}=\frac{3-2}{2.3}+\frac{4-3}{3.4}\frac{5-4}{4.5}+...+\frac{12-11}{11.12}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{12}=\frac{1}{2}-\frac{1}{12}=\frac{5}{12}\)
Nên \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{11^2}>\frac{5}{12}\)
1/2^2+1/3^2+1/4^2+....+1/11^2<1/(2.3)+1/(3.4)+1/(4.5)+.....+1/(11.12)
=1/2-13+1/3-1/4+1/5+.....+1/11-1/12
=1/2-1/12=5/12
VẬY A<5/12
ks cho mình nhé
\(=-\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{100^2}\right)\)
\(=-\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}...\frac{100^2-1}{100^2}\)
\(=-\frac{1.3}{2^2}.\frac{2.4}{3^2}.....\frac{99.101}{100^2}\)
\(=-\frac{1.2....99}{2.3...100}.\frac{3.4....101}{2.3...100}\)
\(=-\frac{1}{100}.\frac{101}{2}=\frac{-101}{200}\)
Học good
\(=-\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{100^2}\right)\)
\(=-\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}...\frac{100^2-1}{100^2}\)
\(=-\frac{1.3}{2^2}\cdot\frac{2.4}{3^2}...\frac{99.101}{100^2}\)
\(=-\frac{1.2...99}{2.3...100}\cdot\frac{3.4...101}{2.3.100}\)
\(=-\frac{1}{100}\cdot\frac{101}{2}\)
\(=-\frac{101}{200}\)
Mk giải ko chép lại đề nhá!
Bài 3:
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}\)\(-\frac{1}{50}\)
\(=\frac{1}{1}-\frac{1}{50}\)
\(=\frac{50}{50}-\frac{1}{50}\)
\(=\frac{49}{50}\)
Vậy: M < 1
Bài 2:
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2013}-\frac{1}{2015}\)
\(=\frac{1}{1}-\frac{1}{2015}\)
\(=\frac{2015}{2015}-\frac{1}{2015}\)
\(=\frac{2014}{2015}\)
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{11^2}>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{11.12}\)
mà \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{11.12}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{2}-\frac{1}{12}=\frac{5}{12}\)
=>\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{11^2}>\frac{5}{12}\)