\(x+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}=\frac{47}{42}\)
tìm x
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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 \(x=2\)
a)Ta đặt A=10+15+...+300
Số số hạng của A là:(300-10):5+1=59(số)
Tổng của A là:(10+300).59:2=9145
=>9145+x=6750
=>x=6750-9145
=>x=-2395
b)\(\frac{1}{42}+\frac{1}{30}+\frac{1}{20}+\frac{1}{12}+\frac{1}{6}+\frac{1}{2}-\frac{1}{x+1}=\frac{59}{77}\)
<=>\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{6.7}-\frac{1}{x+1}=\frac{59}{77}\)
<=>\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{6}-\frac{1}{7}-\frac{1}{x+1}=\frac{59}{77}\)
<=>\(1-\frac{1}{7}-\frac{1}{x+1}=\frac{59}{77}\)
<=>\(\frac{6}{7}-\frac{1}{x+1}=\frac{56}{77}\)
<=>\(\frac{1}{x+1}=\frac{6}{7}-\frac{56}{77}=\frac{66}{77}-\frac{56}{77}\)
<=>\(\frac{1}{x+1}=\frac{10}{77}\)
<=>10(x+1)=77
<=>10x+10=77
<=>10x=67
<=>x=6,7
\(\Rightarrow\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{42}\cdot\frac{x}{3}=\frac{5}{21}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{42}\cdot\frac{x}{3}=\frac{5}{21}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{6}+\frac{1}{42}\cdot\frac{x}{3}=\frac{5}{21}\)
\(\Rightarrow\frac{1}{3}+\frac{1}{42}\cdot\frac{x}{3}=\frac{5}{21}\)
\(\Rightarrow\frac{1}{42}\cdot\frac{x}{3}=\frac{5}{21}-\frac{1}{3}\)
\(\Rightarrow\frac{1}{42}\cdot\frac{x}{3}=\frac{-2}{21}\)
\(\Rightarrow\frac{x}{3}=\frac{-2}{21}\div\frac{1}{42}\)
\(\Rightarrow\frac{x}{3}=-4\)
\(\Rightarrow\frac{x}{3}=\frac{-12}{3}\)
\(\Rightarrow x=-12\)
a, 10+15+20+....+295+x.300+x=67
10+15+20+...+295+x(300+1)=67
10+15+20+...+295+x.301=67
8845+x.301=67
67-8845=x.301
-8878=x.301
x=-29/149/301
b,
\(\frac{1}{7.6}+\frac{1}{6.5}+\frac{1}{5.4}+\frac{1}{4.3}+\frac{1}{3.2}+\frac{1}{2.1}-\frac{1}{x+1}=\frac{59}{77}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}-\frac{1}{x+1}=\frac{59}{77}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}-\frac{1}{x+1}=\frac{59}{77}\)\(1-\frac{1}{7}-\frac{1}{x+1}=\frac{59}{77}\)
\(\frac{6}{7}-\frac{1}{x+1}=\frac{59}{77}\)
\(\frac{1}{x+1}=\frac{6}{7}-\frac{59}{77}\)
\(\frac{1}{x+1}=\frac{1}{11}\)
suy ra x+1=11
suy ra x=10
\(\frac{1}{3}-\left(\frac{1}{3.4}-\frac{1}{4.5}-...-\frac{1}{7.8}\right)=x-\frac{5}{18}\)
\(x-\frac{5}{18}=\frac{1}{3}-\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{8}\right)\)
\(x-\frac{5}{18}=\frac{1}{3}-\frac{1}{3}+\frac{1}{8}\)
\(x-\frac{5}{18}=0+\frac{1}{8}\)
\(x-\frac{5}{18}=\frac{1}{8}\)
\(x=\frac{1}{8}+\frac{5}{18}\)
\(x=\frac{9}{72}+\frac{20}{72}\)
\(x=\frac{29}{72}\)
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
\(\left(x+50\%\right):\frac{7}{8}=\frac{5}{7}\)
\(\Rightarrow\left(x+\frac{1}{2}\right)=\frac{5}{7}.\frac{7}{8}\)
\(\Rightarrow x+\frac{1}{2}=\frac{5}{8}\)
\(\Rightarrow x=\frac{5}{8}-\frac{1}{2}\)
\(\Rightarrow x=\frac{1}{8}\)
Vậy...
Mình làm tiếp bài của bạn " I have a crazy idea "
b) \(\frac{25-x}{3}=\frac{15}{2}\)
Áp dụng tỉ lệ thức:
\(\left(25-x\right).2=15.3\)
\(\Rightarrow25-x=\frac{15.3}{2}=\frac{45}{2}\Leftrightarrow x=25-\frac{45}{2}=\frac{5}{2}\)
c) \(x-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}=1\)
\(\Rightarrow x-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)=1\)
\(\Rightarrow x-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)=1\)
\(\Rightarrow x-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{6}-\frac{1}{7}\right)=1\)
\(\Rightarrow x-\left(\frac{1}{1}-\frac{1}{7}\right)=1\Leftrightarrow x-\frac{6}{7}=1\Leftrightarrow x=1+\frac{6}{7}=\frac{13}{7}\)
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 \(x+\frac{5}{6}=\frac{47}{42}\)
\(x=\frac{47}{42}-\frac{5}{6}\)
\(x=\frac{2}{7}\)
\(x+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}=\frac{47}{42}\)
\(x=\frac{47}{42}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\right)\)
\(x=\frac{47}{42}-\frac{5}{6}\)
\(x=\frac{2}{7}.\)