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=> \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}=\frac{2010}{2011}\)
=> \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2010}{2011}\)
=>\(1-\frac{1}{x+1}=\frac{2010}{2011}\)
=> \(\frac{1}{x+1}=\frac{2011}{2011}-\frac{2010}{2011}=\frac{1}{2011}\)
=> x + 1 = 2011
=> x = 2010
\(\left(1-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}\right)\cdot X=\frac{11}{6}\)
\(< =>\left(\frac{1}{2}-\frac{1}{12}-\frac{1}{60}\right)\cdot X=\frac{11}{6}\)
\(< =>\left(\frac{30}{60}-\frac{5}{60}-\frac{1}{60}\right)\cdot X=\frac{11}{6}\)
\(< =>\left(\frac{30-5-1}{60}\right)\cdot X=\frac{11}{6}\)
\(< =>\frac{2}{5}\cdot X=\frac{11}{6}\)
\(< =>X=\frac{11}{6}:\frac{2}{5}\)
\(< =>X=\frac{55}{12}\)
CHUC BAN HOC TOT >.<
\(X-\frac{2}{3}=\frac{1}{2}+\frac{1}{6}+...+\frac{1}{9900}\)
\(=>X-\frac{2}{3}=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(=>X-\frac{2}{3}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=>X-\frac{2}{3}=1-\frac{1}{100}\)
\(=>X-\frac{2}{3}=\frac{100}{100}-\frac{1}{100}\)
\(=>X-\frac{2}{3}=\frac{99}{100}\)
\(=>X=\frac{99}{100}+\frac{2}{3}\)
\(=>X=\frac{497}{300}\)
Lưu ý: dấu chấm thay dấu nhân
\(x-\frac{2}{3}=\frac{1}{2}+\frac{1}{6}+...+\frac{1}{9900}\)
Tổng vế phải gồm : \(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{9900}\)
\(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{9900}=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{98}-\frac{1}{99}\right)+\left(\frac{1}{99}-\frac{1}{100}\right)\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{100}{100}-\frac{1}{100}\)
\(=\frac{99}{100}\)
Với vế trái, ta có : \(x-\frac{2}{3}=\frac{99}{100}\)
\(x-\frac{2}{3}=\frac{99}{100}\)
\(x=\frac{99}{100}+\frac{2}{3}\)
\(x=\frac{497}{300}\)
Cái này lớp 6 :
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+......+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}\)
\(\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}\)
\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{x\left(x+1\right)}\right)=\frac{2011}{2013}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)
<=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{4026}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2011}{4026}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{2}{4026}=\frac{1}{2013}\)
\(\Leftrightarrow x+1=2013\)
=> x = 2012
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}\)
\(\Rightarrow2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2011}{2013}\)
\(\Rightarrow2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2011}{2013}\)
\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)
\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)
\(\Rightarrow1-\frac{2}{x+1}=\frac{2011}{2013}\)
\(\Rightarrow\frac{2}{x+1}=1-\frac{2011}{2013}\)
\(\Rightarrow\frac{2}{x+1}=\frac{2}{2013}\)
\(\Rightarrow x+1=2013\)
\(\Rightarrow x=2013-1\)
\(\Rightarrow x=2012\)
Vậy \(x=2012\)
~ Ủng hộ nhé
\(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)\div x=\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{32}\right)\)
\(\left(\frac{8}{16}+\frac{4}{16}+\frac{2}{16}+\frac{1}{16}\right)\div x=\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{11.12}\right)\)
\(\frac{15}{16}\div x=\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{12}\right)\)
\(\frac{15}{16}\div x=\left(\frac{1}{1}-\frac{1}{12}\right)\)
\(\frac{15}{16}\div x=\frac{11}{12}\)
\(x=\frac{15}{16}\div\frac{11}{12}\)
\(x=\frac{15}{16}\times\frac{12}{11}\)
\(\Rightarrow x=\frac{180}{176}=\frac{45}{44}\)
a) \(\frac{1,11+0,19-12,2}{0,296+0,094}-x=\left(\frac{1}{2}+\frac{1}{3}\right):2\)
Đề đúng:
\(\frac{1,11+0,19-1,3.2^6}{0,296+0,094}-x=\left(\frac{1}{2}+\frac{1}{3}\right):2\)
\(\frac{1,3-1,3.64}{0,39}-x=\left(\frac{3}{6}+\frac{2}{6}\right).\frac{1}{2}\)
\(\frac{1,3.\left(-63\right)}{0,39}-x=\frac{5}{6}.\frac{1}{2}\)
\(\frac{-81,9}{0,39}-x=\frac{5}{12}\)
\(\left(-210\right)-x=\frac{5}{12}\)
\(x=\left(-210\right)-\frac{5}{12}\)
\(x=\frac{-2520}{12}-\frac{5}{12}\)
\(x=\frac{-2525}{12}\)
b) \(x:\left(3\frac{1}{2}-5\frac{1}{6}\right)=4\frac{1}{5}-6\frac{2}{3}\)
\(x:\left(\frac{7}{2}-\frac{31}{6}\right)=\frac{21}{5}-\frac{20}{3}\)
\(x:\left(\frac{21}{6}-\frac{31}{6}\right)=\frac{63}{15}-\frac{100}{15}\)
\(x:\frac{-10}{6}=\frac{-37}{15}\)
\(x.\frac{6}{-10}=\frac{-37}{5}\)
\(x.\frac{-6}{10}=\frac{-37}{5}\)
\(x.\frac{-3}{5}=\frac{-37}{5}\)
\(x=\frac{-37}{5}:\frac{-3}{5}\)
\(x=\frac{-37}{5}.\frac{5}{-3}\)
\(x=\frac{-37}{-3}\)
\(x=\frac{37}{3}\)
Sai đề bài
Lẽ ra câu a chỉ chia 2 thì chia 22
Sai đề bài mina-san
Ta có \(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right):x=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{132}\)
\(\frac{15}{16}:x=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{11.12}\)
\(\frac{15}{16}:x=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{12}\)
\(\frac{15}{16}:x=1-\frac{1}{12}\)
\(\frac{15}{16}:x=\frac{11}{12}\)
\(x=\frac{15}{16}:\frac{11}{12}\)
\(x=\frac{180}{176}\)
Đúng thì like nha
x = 2 -1/2
x = 6/4
6/4 + 2/4 = 2
ta có: 2 : 2/4 = 4/1 = 4
xong ra được 6/4 + 2/4 = 8/4 = 2