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VM
0
29 tháng 7 2015
=> \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}=\frac{2010}{2011}\)
=> \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2010}{2011}\)
=>\(1-\frac{1}{x+1}=\frac{2010}{2011}\)
=> \(\frac{1}{x+1}=\frac{2011}{2011}-\frac{2010}{2011}=\frac{1}{2011}\)
=> x + 1 = 2011
=> x = 2010
VN
3
HT
26 tháng 9 2020
\(x+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}=\frac{47}{42}\)
\(x+\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\right)=\frac{47}{42}\)
\(x+A=\frac{47}{42}\)
ta thấy :
\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\)
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(A=\frac{1}{1}-\frac{1}{6}\)
\(A=\frac{5}{6}\)
vậy \(x+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}=\frac{47}{42}\)
hay \(x+\frac{5}{6}=\frac{47}{42}\)
\(x=\frac{47}{42}-\frac{5}{6}\)
\(x=\frac{2}{7}\)
26 tháng 9 2020
\(x+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}=\frac{47}{42}\)
\(x=\frac{47}{42}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\right)\)
\(x=\frac{47}{42}-\frac{5}{6}\)
\(x=\frac{2}{7}.\)
11 tháng 3 2020
\(1\frac{1}{12}:\left(\frac{1}{3}-\frac{1}{4}\right)-\frac{2}{x}=\frac{2}{5}:\left(\frac{1}{2}-\frac{1}{5}\right)\)
\(\frac{13}{12}:\frac{1}{12}-\frac{2}{x}=\frac{2}{5}:\frac{3}{10}\)
\(13-\frac{2}{x}=\frac{4}{3}\)
\(\frac{2}{x}=\frac{35}{3}\)
\(6=35x\)
\(x=\frac{6}{35}\)
DP
23 tháng 6 2017
\(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)\div x=\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{32}\right)\)
\(\left(\frac{8}{16}+\frac{4}{16}+\frac{2}{16}+\frac{1}{16}\right)\div x=\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{11.12}\right)\)
\(\frac{15}{16}\div x=\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{12}\right)\)
\(\frac{15}{16}\div x=\left(\frac{1}{1}-\frac{1}{12}\right)\)
\(\frac{15}{16}\div x=\frac{11}{12}\)
\(x=\frac{15}{16}\div\frac{11}{12}\)
\(x=\frac{15}{16}\times\frac{12}{11}\)
\(\Rightarrow x=\frac{180}{176}=\frac{45}{44}\)
18 tháng 4 2015
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
12 tháng 8 2019
\(\left(1-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}\right)\cdot X=\frac{11}{6}\)
\(< =>\left(\frac{1}{2}-\frac{1}{12}-\frac{1}{60}\right)\cdot X=\frac{11}{6}\)
\(< =>\left(\frac{30}{60}-\frac{5}{60}-\frac{1}{60}\right)\cdot X=\frac{11}{6}\)
\(< =>\left(\frac{30-5-1}{60}\right)\cdot X=\frac{11}{6}\)
\(< =>\frac{2}{5}\cdot X=\frac{11}{6}\)
\(< =>X=\frac{11}{6}:\frac{2}{5}\)
\(< =>X=\frac{55}{12}\)
CHUC BAN HOC TOT >.<
KS
3
3 tháng 8 2017
\(\frac{4}{3}+x.\frac{2}{3}-\frac{1}{4}=\frac{53}{12}\)
\(x.\frac{2}{3}-\frac{1}{4}=\frac{53}{12}-\frac{4}{3}\)
\(x.\frac{2}{3}-\frac{1}{4}=\frac{37}{12}\)
\(x.\frac{2}{3}=\frac{37}{12}+\frac{1}{4}\)
\(x.\frac{2}{3}=\frac{10}{3}\)
\(x=\frac{10}{3}:\frac{2}{3}\)
\(x=\frac{30}{6}=5\)
Vậy x = 5
mik ko chắc
3 tháng 8 2017
\(\frac{4}{3}+x.\frac{2}{3}-\frac{1}{4}=4\frac{5}{12}\)
\(\frac{4}{3}+x.\frac{2}{3}=4\frac{5}{12}+\frac{1}{4}\)
\(\frac{4}{3}+x.\frac{2}{3}=\frac{53}{12}+\frac{1}{4}\)
\(\frac{4}{3}+x.\frac{2}{3}=\frac{14}{3}\)
\(x.\frac{2}{3}=\frac{14}{3}-\frac{4}{3}\)
\(x.\frac{2}{3}=\frac{10}{3}\)
\(x=\frac{10}{3}:\frac{2}{3}\)
\(x=5\)
<=> \(\frac{12\left(x+2\right)}{x\left(x+2\right)}\)= \(\frac{12x}{\left(x+2\right)x}\)+ \(\frac{\left(x+2\right)x}{\left(x+2\right)x}\)
<=> 12x + 24 = 12x + x2 + 2x
<=> 12x - 12x -x2 + 2x + 24 = 0
<=> - x2 + 2x + 24 = 0
<=> - x2 - 4x + 6x + 24 = 0
<=> - x(x - 4) + 6(x + 4) = 0
<=> (x + 4)(- x + 6) = 0
<=> x + 4 = 0 hoặc - x + 6 = 0
<=> x = -4 hoặc x = - 6
Cho tam giác ABC trên cạnh AB lấy điểm D sao cho AD gấp đôi DB trên AC lấy điểm E sao cho AE gấp đôi EC , BE cắt CD tại G .So sánh diện tích GDB với điện tích GEC