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BM
2
LT
17 tháng 10 2017
\(\left(\frac{1}{5}+\frac{1}{35}+\frac{1}{63}\right)\cdot x=1\)
\(\frac{11}{45}\cdot x=1\)
\(x=\frac{45}{11}\)
17 tháng 10 2017
(1/5+1/35+1/63)*x=1
(441/2205+63/2205+35/2205)*x=1
11/45*x=1
x=1:11/45
x=45/11
TT
2
26 tháng 8 2020
\(\left(\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)×x=\frac{6}{5}\)
\(\frac{3}{55}×x=\frac{6}{5}\)
\(x=\frac{6}{5}:\frac{3}{55}\)
\(x=22\)
vậy \(x=22\)
VM
26 tháng 8 2020
\(\left(\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right).x=\frac{6}{5}\)
\(\Leftrightarrow\left(\frac{1.99}{35.99}+\frac{1.55}{63.55}+\frac{1.35}{99.35}\right).x=\frac{6}{5}\)
\(\Leftrightarrow\left(\frac{99}{3465}+\frac{55}{3465}+\frac{35}{3465}\right).x=\frac{6}{5}\)
\(\Leftrightarrow\left(\frac{99+55+35}{3465}\right).x=\frac{6}{5}\)
\(\Leftrightarrow\frac{189}{3465}.x=\frac{6}{5}\)
\(\Leftrightarrow\frac{3}{55}.x=\frac{6}{5}\)
\(\Leftrightarrow x=\frac{6}{5}:\frac{3}{55}\)
\(\Leftrightarrow x=\frac{6}{5}.\frac{55}{3}\)
\(\Leftrightarrow x=\frac{330}{15}\)
\(\Leftrightarrow x=22\)
Vậy \(x=22.\)
PT
1
CM
12 tháng 1 2018
1/3x5+1/5x7+1/7x9)xX=1
1/2x(1/3-1/5+1/5-1/7+1/7-1/9)xX=1
1/2x(1/3-1/9)xX=1
1/2x2/9xX=1
1/9xX=1
X=9
PT
1
CM
3 tháng 5 2018
1/3x5+1/5x7+1/7x9)xX=11/2x(1/3-1/5+1/5-1/7+1/7-1/9)xX=11/2x(1/3-1/9)xX=11/2x2/9xX=11/9xX=1X=9
23 tháng 8 2016
\(\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right).y=\frac{2}{3}\)
\(\frac{1}{2}.\left(1-\frac{1}{3}\right)+\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}\right)+\frac{1}{2}.\left(\frac{1}{7}-\frac{1}{9}\right)+\frac{1}{2}.\left(\frac{1}{9}-\frac{1}{11}\right).y=\frac{2}{3}\)
\(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right).y=\frac{2}{3}\)
\(\frac{1}{2}.\left(1-\frac{1}{11}\right).y=\frac{2}{3}\)
\(\left(1-\frac{1}{11}\right).y=\frac{4}{3}\)
\(\frac{10}{11}.y=\frac{4}{3}\)
\(\Rightarrow y=\frac{22}{15}\)
S
2
NH
30 tháng 3 2019
\((\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99})x=\frac{2}{3}\)
Đặt \(A=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\)
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
\(A=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{9.11}\right)\)
\(A=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\right)\)
\(A=\frac{1}{2}\left(1-\frac{1}{11}\right)\)
\(A=\frac{1}{2}.\frac{10}{11}=\frac{5}{11}\)
Thay A vào biểu thức
\(\Rightarrow\frac{5}{11}x=\frac{2}{3}\)
\(\Rightarrow x=\frac{22}{15}\)
P/s: Có thể tính sai :(
30 tháng 3 2019
\(\left[\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right]\times x=\frac{2}{3}\)
Trước tiên mình tính dãy có dấu ngoặc đã
Đặt : \(S=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\)
\(=\frac{1}{2}\left[\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}\right]\)
\(=\frac{1}{2}\left[\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}\right]\)
\(=\frac{1}{2}\left[1-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{11}\right]\)
\(=\frac{1}{2}\left[1-\frac{1}{11}\right]=\frac{1}{2}\cdot\frac{10}{11}=\frac{1\cdot10}{2\cdot11}=\frac{1\cdot5}{1\cdot11}=\frac{5}{11}\)
Thay vào biểu thức \(S=\frac{5}{11}\)ta lại có :
\(\frac{5}{11}\times x=\frac{2}{3}\)
\(\Leftrightarrow x=\frac{2}{3}:\frac{5}{11}\)
\(\Leftrightarrow x=\frac{2}{3}\cdot\frac{11}{5}\)
\(\Leftrightarrow x=\frac{22}{15}\)
Vậy \(x=\frac{22}{15}\)
[ 1/5 + 1/35 + 1/63 ] x [ x-2 ] =1
\(\frac{63+9+5}{315}x\left(x-2\right)=1\)
\(\frac{77}{315}x\left(x-2\right)=1\)
\(\frac{11}{45}x\left(x-2\right)=1\)
\(x-2=1:\frac{11}{45}=\frac{45}{11}\)
\(x=\frac{45}{11}+2=\frac{67}{11}\)
\(\left(\frac{1}{5}+\frac{1}{35}+\frac{1}{63}\right)\times\left(x-2\right)=1\)
\(\frac{11}{45}\times\left(x-2\right)=1\)
\(x-2=\frac{45}{11}\)
\(x=\frac{67}{11}\)