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MD
31 tháng 8 2017
\(a,\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{x.\left(x+3\right)}=\dfrac{101}{1540}\)
\(\dfrac{1}{3}.3.\left[\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{x.\left(x+3\right)}\right]=\dfrac{101}{1540}\)
\(\dfrac{1}{3}.\left[\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+...+\dfrac{3}{x.\left(x+3\right)}\right]=\dfrac{101}{1540}\)
\(\dfrac{1}{3}.\left[\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right]=\dfrac{101}{1540}\)
\(\dfrac{1}{3}.\left(\dfrac{1}{5-1}-\dfrac{1}{x+3}\right)=\dfrac{101}{1540}\)
\(\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{101}{1540}.\dfrac{1}{3}\)
\(\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)
\(\dfrac{1}{x+3}=\dfrac{1}{3}-\dfrac{303}{1540}\)
\(\dfrac{1}{x+3}=\dfrac{1}{308}\)
\(\Rightarrow x+3=308\)
\(x=308-3\)
\(x=305\)
\(b,1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{x.\left(x+1\right):2}=1\dfrac{1991}{1993}\)
\(\dfrac{1}{2}.\left(1+\dfrac{1}{3}+\dfrac{1}{6}+...+\dfrac{1}{x.\left(x+1\right):2}\right)=\dfrac{3984}{3986}\)
\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{8}+...+\dfrac{1}{x.\left(x+1\right)}=\dfrac{3984}{3986}\)
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{x.\left(x+1\right)}=\dfrac{3984}{3986}\)
\(\dfrac{2-1}{1.2}+\dfrac{4-3}{3.4}+...+x+1-\dfrac{x}{x.\left(x+1\right)}=\dfrac{3984}{3986}\)
\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}+\dfrac{1}{x+1}=\dfrac{3984}{3986}\)
\(1-\dfrac{1}{x+1}=\dfrac{3984}{3986}\)
\(\dfrac{1}{x+1}=1-\dfrac{3984}{3986}\)
\(\dfrac{1}{x+1}=\dfrac{1}{1993}\)
=>\(x+1=1993\)
\(x=1993-1\)
\(x=1992\)
AN
1 tháng 9 2017
A= \(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{4.5.6}+....+\dfrac{1}{37.38.39}\)
A=\(\dfrac{1}{1}-\dfrac{1}{39}\)
A=\(\dfrac{38}{39}\)
còn lại tự làm do mình có việc chút
23 tháng 2 2019
\(\frac{x-4}{y-3}=\frac{4}{3}\)
\(\Rightarrow\left(x-4\right).3=\left(y-3\right).4\)
\(3x-12=4y-12\)
\(\Leftrightarrow3x=4y\)
\(\Rightarrow\frac{x}{\frac{1}{3}}=\frac{y}{\frac{1}{4}}\)
Áp dụng tính chất của dãy tỉ số bằng nhau , ta có:
\(\frac{x}{\frac{1}{3}}=\frac{y}{\frac{1}{4}}=\frac{x-y}{\frac{1}{3}-\frac{1}{4}}=\frac{5}{\frac{1}{12}}=5.12=60\)
\(\Rightarrow\hept{\begin{cases}x=60.\frac{1}{3}=20\\y=60.\frac{1}{4}=15\end{cases}}\)
Vậy x = 20 ; y = 15
21 tháng 3 2017
a)\(\frac{5}{2}-3\left(\frac{1}{3}-x\right)=\frac{1}{4}-7x\)
\(\Leftrightarrow\frac{5}{2}-1+x=\frac{1}{4}-7x\)
\(\Leftrightarrow8x=-\frac{5}{4}\)
\(\Leftrightarrow x=-\frac{5}{32}\)
c)\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)
\(\Leftrightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2003}\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2003}\)
\(\Leftrightarrow x+1=2003\)
\(\Leftrightarrow x=2002\)
31 tháng 8 2017
a)
\(A=\dfrac{2}{3.4}+\dfrac{2}{4.5}+\dfrac{2}{5.6}+...+\dfrac{2}{52.53}\\ A=2\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{53.54}\right)\\ A=2.\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{52}-\dfrac{1}{53}\right)\\ A=2\left(\dfrac{1}{3}-\dfrac{1}{53}\right)\\ A=\dfrac{100}{3.53}=\dfrac{100}{159}\)
b)
\(B=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{2652}\\ B=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{51.52}\\ B=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{51}-\dfrac{1}{52}\\ B=\dfrac{1}{2}-\dfrac{1}{52}=\dfrac{50}{104}=\dfrac{25}{52}\)
câu c tương tự câu a
2 tháng 7 2022
\(\Leftrightarrow2\left(\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{1005}{1006}\)
\(\Leftrightarrow\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{1005}{2012}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{-251}{1006}\)
=>x+1=-1006/251
hay x=-1257/251
15 tháng 3 2018
\(2\dfrac{x}{7}=\dfrac{14+x}{7}=\dfrac{75}{35}=\dfrac{15}{7}\\ \Rightarrow\left(14+x\right).7=15.7\\ 14+x=15\\ x=15-14=1\)
25 tháng 8 2017
a) x.(\(\dfrac{6}{7}\)+\(\dfrac{5}{6}\))=\(\dfrac{3}{4}\)
x.\(\dfrac{71}{42}\)=\(\dfrac{3}{4}\)
x=\(\dfrac{3}{4}\):\(\dfrac{71}{42}\)
x=\(\dfrac{63}{142}\)
TT
25 tháng 8 2017
a.\(\dfrac{6}{7}x+\dfrac{5}{6}x=\dfrac{3}{4}\)
\(x.\left(\dfrac{6}{7}+\dfrac{5}{6}\right)=\dfrac{3}{4}\)
\(x.\dfrac{71}{42}=\dfrac{3}{4}\)
\(x=\dfrac{3}{4}:\dfrac{71}{42}\)
\(x=\dfrac{63}{142}\)
b\(\dfrac{5}{4}-\dfrac{3}{5}:x=1\dfrac{1}{3}\)
\(\dfrac{3}{5}:x=\dfrac{5}{4}-1\dfrac{1}{3}\)
\(\dfrac{3}{5}:x=\dfrac{-1}{12}\)
\(x=\dfrac{3}{5}:\dfrac{-1}{12}\)
\(x=\dfrac{-36}{5}\)
c. \(\left(\dfrac{4}{7}x-\dfrac{1}{3}\right):3\dfrac{1}{2}=0,5\)
\(\left(\dfrac{4}{7}x-\dfrac{1}{3}\right)=0,5:3\dfrac{1}{2}\)
\(\dfrac{4}{7}x-\dfrac{1}{3}=\dfrac{1}{7}\)
\(\dfrac{4}{7}x=\dfrac{1}{7}+\dfrac{1}{3}\)
\(\dfrac{4}{7}x=\dfrac{10}{21}\)
\(x=\dfrac{10}{21}:\dfrac{4}{7}\)
\(x=\dfrac{5}{6}\)
d.\(\dfrac{4}{5}-\dfrac{2}{3}x=1\dfrac{1}{4}+2,5x\)
\(\dfrac{4}{5}-\left(\dfrac{2}{3}x-2,5x\right)=1\dfrac{1}{4}\)
\(\dfrac{4}{5}-\dfrac{-11}{6}x=1\dfrac{1}{4}\)
\(\dfrac{-11}{6}x=\dfrac{4}{5}-1\dfrac{1}{4}\)
\(\dfrac{-11}{6}x=\dfrac{-9}{20}\)
\(x=\dfrac{-9}{20}:\dfrac{-11}{6}\)
\(x=\dfrac{27}{110}\)
có sai sót j xin bn thông cảm !
13 tháng 8 2017
a)\(\dfrac{2}{3}x-\dfrac{5}{6}=1\dfrac{1}{4}\)
\(\dfrac{2}{3}x-\dfrac{5}{6}=\dfrac{5}{4}\)
\(\dfrac{2}{3}x=\dfrac{5}{4}+\dfrac{5}{6}\)
\(\dfrac{2}{3}x=\dfrac{25}{12}\)
\(x=\dfrac{25}{12}:\dfrac{2}{3}\)
=>\(x=\dfrac{25}{8}\)
13 tháng 8 2017
a) \(\dfrac{2}{3}x-\dfrac{5}{6}=1\dfrac{1}{4}\) b) \(2\dfrac{1}{3}-\dfrac{4}{5}:x=0,2\)
\(\dfrac{2}{3}x-\dfrac{5}{6}=\dfrac{5}{4}\) \(\dfrac{7}{3}-\dfrac{4}{5}:x=\dfrac{1}{5}\)
\(\dfrac{2}{3}x=\dfrac{5}{4}-\dfrac{5}{6}\) \(\dfrac{4}{5}:x=\dfrac{7}{3}-\dfrac{1}{5}\)
\(\dfrac{2}{3}x=\dfrac{30}{24}-\dfrac{20}{24}\) \(\dfrac{4}{5}:x=\dfrac{35}{15}-\dfrac{3}{15}\)
\(\dfrac{2}{3}x=\dfrac{5}{12}\) \(\dfrac{4}{5}:x=\dfrac{32}{15}\)
\(x=\dfrac{5}{12}:\dfrac{2}{3}\) \(x=\dfrac{4}{5}:\dfrac{32}{15}\)
\(x=\dfrac{5}{12}:\dfrac{8}{12}\) \(x=\dfrac{4}{5}.\dfrac{15}{32}\)
\(x=\dfrac{5}{12}.\dfrac{12}{8}=\dfrac{5}{8}\) \(x=\dfrac{4.15}{5.32}\)
\(x=\dfrac{1.3}{1.8}=\dfrac{3}{8}\)
d)\(\left(\dfrac{4}{3}-\dfrac{1}{4}x\right)^3=\dfrac{-8}{27}\)
\(\left(\dfrac{4}{3}-\dfrac{1}{4}x\right)^3=\left(\dfrac{-2}{3}\right)^3\)
\(\Rightarrow\dfrac{4}{3}-\dfrac{1}{4}x=\dfrac{-2}{3}\)
\(\Rightarrow\dfrac{1}{4}x=\dfrac{4}{3}-\dfrac{-2}{3}\)
\(\Rightarrow\dfrac{1}{4}x=2\)
\(\Rightarrow x=2:\dfrac{1}{4}\)
\(\Rightarrow x=2.4=8\)
ai trả lời nhanh mik tick
\(\dfrac{x}{3}=-\dfrac{4}{6}\)
=>\(\dfrac{x}{3}=-\dfrac{2}{3}\)
=>x=-2