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\(1\frac{13}{15}.0,75-\left(\frac{8}{15}+25\%\right).\frac{24}{47}-3\frac{12}{13}:3\)
\(=\frac{28}{15}.\frac{3}{4}-\left(\frac{8}{15}+\frac{1}{4}\right).\frac{24}{47}-\frac{51}{13}:3\)
\(=\frac{7}{5}-\frac{47}{60}.\frac{24}{47}-\frac{17}{13}\)
\(=\frac{7}{5}-\frac{2}{5}-\frac{17}{13}\)
\(=\frac{-4}{13}\)
\(4\frac{1}{3}.\left(\frac{1}{6}-\frac{1}{2}\right)\le x\le\frac{2}{3}.\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)
\(\Leftrightarrow\frac{13}{3}.\frac{-1}{3}\le x\le\frac{2}{3}.\frac{-11}{12}\)
\(\Leftrightarrow\frac{-13}{9}\le x\le\frac{-11}{18}\)
\(\Leftrightarrow x=-1\)
Tìm x
\(a,2x-25\%=\frac{1}{2}\)
\(b,\left(\frac{3x}{7}+1\right).\left(-0,25\right)=\frac{-1}{28}\)
\(\)
mik ko ghi lại đề nhé!
\(A=\left(\frac{18}{15}.\frac{1}{4}.3\right)+\left(-\frac{47}{60}\right).\frac{24}{47}\)
\(A=\frac{8}{5}+\left(-\frac{2}{5}\right)\)
\(A=\frac{6}{5}\)
\(B=\frac{3}{4}.\frac{28}{15}-\left(\frac{8}{15}+\frac{1}{4}\right).\frac{24}{47}-\frac{17}{13}\)
\(B=\frac{7}{5}-\frac{47}{60}.\frac{24}{47}-\frac{17}{13}\)
\(B=\frac{7}{5}-\frac{2}{5}-\frac{17}{13}\)
\(B=1-\frac{17}{13}\)
\(B=-\frac{4}{13}\)
THANKS
TÌM X
a,\(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
b, \(\left(x-\frac{1}{2}\right)^2=\frac{4}{25}\)
\(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}=\left(\frac{1}{3}\right)^3\)
\(\Leftrightarrow x-\frac{1}{2}=\frac{1}{3}\)
\(\Leftrightarrow x=\frac{1}{3}+\frac{1}{2}\)
\(\Leftrightarrow x=\frac{5}{6}\)
Bài làm
a) \(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\) b) \(\left(x-\frac{1}{2}\right)^2=\frac{4}{25}\)
=> \(\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\) => \(\left(x-\frac{1}{2}\right)^2=\left(\frac{2}{5}\right)^2\)
=> \(x-\frac{1}{2}=\frac{1}{3}\) => \(x-\frac{1}{2}=\frac{2}{5}\)
\(x=\frac{1}{3}+\frac{1}{2}\) \(x=\frac{2}{5}+\frac{1}{2}\)
\(x=\frac{2}{6}+\frac{3}{6}\) \(x=\frac{4}{10}+\frac{5}{10}\)
\(x=\frac{5}{6}\) \(x=\frac{9}{10}\)
Vậy \(x=\frac{5}{6}\) Vậy \(x=\frac{9}{10}\)
# Chúc bạn học tốt #
\(A=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)....\left(1-\frac{1}{1+2+...+100}\right)\)
\(A=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right).....\left(1-\frac{1}{5050}\right)\)
\(A=\frac{2}{3}.\frac{5}{6}....\frac{5049}{5050}=\frac{4}{6}.\frac{10}{12}...\frac{10098}{10100}\)
\(A=\frac{1.4}{2.3}.\frac{2.5}{3.4}...\frac{99.102}{100.101}\)
\(A=\frac{1.2...98.99}{2.3...99.100}.\frac{4.5...102}{3.4...101}=\frac{1}{100}.\frac{102}{3}\)
Vậy \(A=\frac{102}{300}=\frac{17}{50}\)
\(A=\left(1-\frac{1}{1+2}\right).\left(1-\frac{1}{1+2+3}\right).....\left(1-\frac{1}{1+2+...+100}\right)\)
\(=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)...\left(1-\frac{1}{5050}\right)\)
\(=\frac{2}{3}.\frac{5}{6}.....\frac{5049}{5050}\)
\(=\frac{4}{6}.\frac{10}{12}.....\frac{10098}{10100}\)
\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.....\frac{99.102}{100.101}\)
\(=\frac{1.2.3..98.99}{2.3.4..99.100}.\frac{4.5.6...102}{3.4.5...101}\)
\(=\frac{1}{100}.\frac{102}{3}\)
\(=\frac{17}{50}\)
Ta có:\(n^2+\left(n+1\right)^2=n^2+n^2+2n+1=2n^2+2n+1>2n^2+2n=2n\left(n+1\right)\)
\(\Rightarrow\frac{1}{n^2+\left(n+1\right)^2}< \frac{1}{2n\left(n+1\right)}\)
Áp dụng vào bài toán,ta có:
\(\frac{1}{1^2+2^2}+\frac{1}{2^2+3^2}+\frac{1}{3^2+4^2}+......+\frac{1}{n^2+\left(n+1\right)^2}\)
\(< \frac{1}{2\cdot1\cdot2}+\frac{1}{2\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+.....+\frac{1}{2\cdot n\cdot\left(n+1\right)}\)
\(=\frac{1}{2}\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{n\left(n+1\right)}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{n}-\frac{1}{n+1}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{n+1}\right)\)
\(=\frac{1}{2}-\frac{1}{2\left(n+1\right)}\)
\(< \frac{1}{2}\)