Tìm x:
(\(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+....+\frac{1}{90}\)) .x = \(\frac{14}{30}\)
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\(\Leftrightarrow\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\times\frac{x}{3}=\frac{5}{21}\)
\(\Leftrightarrow\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\times\frac{x}{3}=\frac{5}{21}\)
\(\Leftrightarrow\left(\frac{1}{2}-\frac{1}{7}\right)\times\frac{x}{3}=\frac{5}{21}\)
\(\Leftrightarrow\frac{5}{14}\times\frac{x}{3}=\frac{5}{21}\)
\(\Leftrightarrow\frac{x}{3}=\frac{2}{3}\)
\(\Leftrightarrow x=2\)
\(\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right).\frac{x}{3}=\frac{5}{21}\)
\(\Rightarrow\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right).\frac{x}{3}=\frac{5}{21}\)
\(\Rightarrow\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}\right).\frac{x}{3}=\frac{5}{21}\)
\(\Rightarrow\left(\frac{1}{2}-\frac{1}{7}\right).\frac{x}{3}=\frac{5}{21}\)
\(\Rightarrow\left(\frac{7}{14}-\frac{2}{14}\right).\frac{x}{3}=\frac{5}{21}\)
\(\Rightarrow\frac{5}{14}.\frac{x}{3}=\frac{5}{21}\)
\(\Rightarrow\frac{x}{3}=\frac{5}{21}:\frac{5}{14}\)
\(\Rightarrow\frac{x}{3}=\frac{2}{3}\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
\(\frac{1}{3}-\left(\frac{1}{3.4}-\frac{1}{4.5}-...-\frac{1}{7.8}\right)=x-\frac{5}{18}\)
\(x-\frac{5}{18}=\frac{1}{3}-\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{8}\right)\)
\(x-\frac{5}{18}=\frac{1}{3}-\frac{1}{3}+\frac{1}{8}\)
\(x-\frac{5}{18}=0+\frac{1}{8}\)
\(x-\frac{5}{18}=\frac{1}{8}\)
\(x=\frac{1}{8}+\frac{5}{18}\)
\(x=\frac{9}{72}+\frac{20}{72}\)
\(x=\frac{29}{72}\)
1/3 - 1/12 - 1/20 - 1/30 - 1/42 - 1/56 = x - 5/18
1/4 - 1/20 - 1/30 - 1/42 - 1/56 = x - 5/18
1/5 - 1/30 - 1/42 - 1/56 = x - 5/18
1/6 - 1/42 - 1/56 = x - 5/18
1/7 - 1/56 = x - 5/18
1/8 = x - 5/18
x=1/8+5/18
x= 29/72
Vậy : x = 29/72
ta có:$\frac{x-1}{12}+\frac{x-1}{20}+\frac{x-1}{30}+\frac{x-1}{42}+\frac{x-1}{56}+\frac{x-1}{72}=\frac{16}{9}$
=> x+1(1/12+1/20+1/30+1/42+1/56+1/72)=16/9
=> x+1.2/9=16/9
=> x+1 = (16/9):(2/9)
=> x+1 = 8
=> x = 9
thông cảm mình ko đánh được dấu ngoặc tròn
[x-1].[1/12+1/20+1/30+1/42+1/56+1/72] =16/9
[x-1].[1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9]=16/9
[x-1].[1/3-1/9]=16/9
[x-1].2/9=16/9
x-1=16/9:2/9
x-1=8
x=7
Vậy x=7
\(x+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}=\frac{47}{42}\)
\(x+\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\right)=\frac{47}{42}\)
\(x+A=\frac{47}{42}\)
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 \(x+\frac{5}{6}=\frac{47}{42}\)
\(x=\frac{47}{42}-\frac{5}{6}\)
\(x=\frac{2}{7}\)
\(x+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}=\frac{47}{42}\)
\(x=\frac{47}{42}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\right)\)
\(x=\frac{47}{42}-\frac{5}{6}\)
\(x=\frac{2}{7}.\)
\(\left(1-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}\right)\cdot X=\frac{11}{6}\)
\(< =>\left(\frac{1}{2}-\frac{1}{12}-\frac{1}{60}\right)\cdot X=\frac{11}{6}\)
\(< =>\left(\frac{30}{60}-\frac{5}{60}-\frac{1}{60}\right)\cdot X=\frac{11}{6}\)
\(< =>\left(\frac{30-5-1}{60}\right)\cdot X=\frac{11}{6}\)
\(< =>\frac{2}{5}\cdot X=\frac{11}{6}\)
\(< =>X=\frac{11}{6}:\frac{2}{5}\)
\(< =>X=\frac{55}{12}\)
CHUC BAN HOC TOT >.<
\(\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{90}\right).x=\frac{14}{30}\)
\(\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{9.10}\right).x=\frac{14}{30}\)
\(\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}\right).x=\frac{14}{30}\)
\(\left(\frac{1}{3}-\frac{1}{10}\right).x=\frac{14}{30}\)
\(\frac{10-3}{30}.x=\frac{14}{30}\)
\(\frac{7}{30}.x=\frac{14}{30}\)
\(x=\frac{14}{30}:\frac{7}{30}\)
\(x=\frac{14}{30}.\frac{30}{7}\)
\(x=2\)
Bn giải chi tiết giúp mình bước thứ 3 được ko