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#)Giải :
a)\(2009^{\left(1000-1^3\right)\left(1000-2^3\right)...\left(1000-15^3\right)}=2009^{\left(1000-1^3\right)...\left(1000-10^3\right)...\left(1000-15^3\right)}=2009^0=1\)
b)\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...\left(\frac{1}{125}-\frac{1}{5^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...0...\left(\frac{1}{125}-\frac{1}{25^3}\right)=0\)
\(\left(\frac{1}{38}-1\right).\left(\frac{1}{37}-1\right).\left(\frac{1}{36}-1\right)....\left(\frac{1}{2}-1\right)\)
\(=-\left(1-\frac{1}{38}\right).\left(1-\frac{1}{37}\right)....\left(1-\frac{1}{2}\right)\)
\(=-\frac{37}{38}.\frac{36}{37}...\frac{1}{2}\)
\(=\frac{-1}{38}\)
\(\left(\frac{1}{38}-1\right)\left(\frac{1}{37}-1\right)\left(\frac{1}{36}-1\right).......\left(\frac{1}{2}-1\right)\)
\(=\frac{-37}{38}.\frac{-36}{37}.\frac{-35}{36}.......\frac{-2}{3}.\frac{-1}{2}\)
\(=\frac{-1}{38}\)
a, \(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{x\left(x+1\right)}=\frac{13}{90}\)
⇒ \(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{13}{90}\)
⇒ \(\frac{1}{5}-\frac{1}{x+1}=\frac{13}{90}\)
⇒ \(\frac{1}{x+1}=\frac{1}{5}-\frac{13}{90}\)
⇒ \(\frac{1}{x+1}=\frac{18}{90}-\frac{13}{90}\)
⇒ \(\frac{1}{x+1}=\frac{1}{18}\)
⇒ x + 1 = 18
⇒ x = 17
Vậy x = 17
b, \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}=\frac{49}{148}\)
⇒ \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x\left(x+3\right)}=\frac{49.3}{148}\)
⇒ \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{147}{148}\)
⇒ \(1-\frac{1}{x+3}=\frac{147}{148}\)
⇒ \(\frac{1}{x+3}=1-\frac{147}{148}\)
⇒ \(\frac{1}{x+3}=\frac{1}{148}\)
⇒ x + 3 = 148
⇒ x = 145
Vậy x = 145
\(A=xemlai\) chưa hưa hiểu Quy luật
\(B=\frac{\left(n.\left(n+2\right)+1\right)}{n\left(n+2\right)}=\frac{\left(n+1\right)^2}{n.\left(n+2\right)}\)
\(B=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}.\frac{5.5}{4.5}...\frac{98.98}{97.99}\frac{99.99}{98.100}\frac{100.100}{99.101}\\\)
\(B=\frac{2.100}{1.101}=\frac{200}{101}\)
a) \(\frac{-1}{24}-\left[\frac{1}{4}-\left(\frac{1}{2}-\frac{7}{8}\right)\right]\)
= \(\frac{-1}{24}-\left[\frac{6}{24}-\left(\frac{12}{24}-\frac{21}{24}\right)\right]\)
= \(\frac{-1}{24}-\left[\frac{6}{24}-\frac{-9}{24}\right]\)
= \(\frac{-1}{24}-\frac{15}{24}\)
= \(\frac{-16}{24}\) = \(\frac{-2}{3}\)
b) \(\left(\frac{5}{7}-\frac{7}{5}\right)-\left[\frac{1}{2}-\left(-\frac{2}{7}-\frac{1}{10}\right)\right]\)
= \(\left(\frac{50}{70}-\frac{98}{70}\right)-\left[\frac{35}{70}-\left(-\frac{20}{70}-\frac{7}{70}\right)\right]\)
= \(\frac{-48}{70}-\left[\frac{35}{70}-\left(-\frac{20}{70}-\frac{7}{70}\right)\right]\)
= \(\frac{-48}{70}-\left[\frac{35}{70}-\frac{-27}{70}\right]\)
= \(\frac{-48}{70}-\frac{62}{70}\)
= \(\frac{-110}{70}=\frac{-11}{7}\)
\(=\left(-\frac{37}{38}\right)\left(-\frac{36}{37}\right)\left(-\frac{35}{36}\right)...\left(-\frac{31}{32}\right)=\left(-\frac{31}{38}\right)\)