Tìm x, biết:
\(\frac{1}{1.4}\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}=\frac{125}{376}\)
K
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7 tháng 9 2016
bạn ơi như là cô giáo cho đề sai rồi kết quả phải là \(\frac{375}{376}\)thì mới giải được
TP
12 tháng 8 2017
Ta có:
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}=\frac{125}{376}\)
\(\Rightarrow\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x.\left(x+3\right)}\right)=\frac{125}{376}\)
\(\Rightarrow\frac{1}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{125}{376}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{x+3}=\frac{125}{376}:\frac{1}{3}=\frac{375}{376}\)
\(\Rightarrow\frac{1}{x+3}=1-\frac{375}{376}=\frac{1}{376}\Leftrightarrow x+3=376\Leftrightarrow x=373\)
BT
3 tháng 3 2017
1) Ta có: A=\(\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x\left(x+3\right)}\right)=\)
=\(\frac{1}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\)
=\(\frac{1}{3}\left(1-\frac{1}{x+3}\right)=\frac{1}{3}.\frac{x+2}{x+3}=\frac{125}{376}\)
<=> \(\frac{x+2}{x+3}=\frac{375}{376}\)<=> 376(x+2)=375(x+3) <=> 376x+752=375x+1125 => X=373
25 tháng 4 2016
1/3.(1-1/4+1/4-1/7+......+1/x-1/(x+3)=6/19
1/3.(1-1/x+3)=6/19
1-1/x+3=6/19:1/3
1-1/x+3=18/19
1/x+3=1-18/19
1/x+3=1/19
=> x+3=19
=>x=19-3
x=16
25 tháng 4 2016
Đặt biểu thức là A, ta có:
3A=\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.11}+...+\frac{3}{x\left(x+3\right)}\)
3A=\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+3}\)
3A=1-\(\frac{1}{x+3}\)
A=\(\frac{1}{3}-\frac{3}{x+3}\)
=>\(\frac{1}{3}-\frac{3}{x+3}\) =\(\frac{6}{19}\) =>x=168
D
8 tháng 3 2016
\(\left(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{97.100}\right)=\frac{0,33x}{2009}\)
\(\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{97.100}\right)=\frac{0,33x}{2009}\)
\(\left(1-\frac{1}{4}+\frac{1}{4}-...-\frac{1}{100}\right)=\frac{0,33x}{2009}\)
\(1-\frac{1}{100}=\frac{0,33x}{2009}\)
\(\frac{99}{100}=\frac{0,33x}{2009}\Rightarrow2009x99=0,33x\times100\)
198891:100:0,33=6027=x
TL
5
13 tháng 5 2016
\(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{x\left(x+3\right)}=\frac{18}{19}\)
\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{18}{19}\)
\(1-\frac{1}{x+3}=\frac{18}{19}\)
...............
TN
13 tháng 5 2016
đặt VT là A ta có:
\(3A=3\left(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{x\left(x+3\right)}\right)=\frac{6}{19}\)
\(3A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+3}\)
\(3A=1-\frac{1}{x+3}\)
\(\left(1-\frac{1}{x+3}\right):3\)
thay A vào VT ta đc\(\left(1-\frac{1}{x+3}\right):3=\frac{6}{19}\)
\(1-\frac{1}{x+3}=\frac{18}{19}\)
\(\frac{1}{x+3}=\frac{1}{19}\)
=>x+3=19
=>x=16
MT
10
18 tháng 4 2019
3. ( 1/1.4 +1/4.7 +1/7.10 +...+ 1/x.(x+3)
3/1.4 +1/4.7+1/7.10 + ...+ 3/ x . (x+3)
1/1 - 1/4 + 1/4 - 1/6 + 1/7 - 1/10 + ...+ 1/x-1/x+3
1/1 - 1/x+3
x+3/x+3 - 1/x+3
x+2/x+3
TN
12 tháng 5 2016
đặt VT là A ta đc:
\(3A=3\left(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{x\left(x+3\right)}\right)\)
\(3A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+3}\)
\(3A=1-\frac{1}{x+3}\)
\(A=\left(1-\frac{1}{x+3}\right):3\)
thay A vào VT ta đc:\(\left(1-\frac{1}{x+3}\right):3=\frac{6}{19}\)
\(1-\frac{1}{x+3}=\frac{18}{19}\)
\(\frac{1}{x+3}=\frac{1}{19}\)
=>x+3=19
=>x=16
9 tháng 11 2016
\(\frac{1}{1.3}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{n\left(n+3\right)}=\frac{2018}{6057}\)
\(\Rightarrow\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{n\left(n+3\right)}\right)=\frac{2018}{6057}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}=\frac{2018}{6057}.3\)
\(\Rightarrow1-\frac{1}{n+3}=\frac{2018}{2019}\)
\(\Rightarrow\frac{1}{n+3}=1-\frac{2018}{2019}\)
\(\Rightarrow\frac{1}{n+3}=\frac{1}{2019}\)
\(\Rightarrow n+3=2019\)
\(\Rightarrow n=2016\)
Vậy n = 2016
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}=\frac{125}{376}\)
\(3.\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}\right)=3.\frac{125}{376}\)
\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x\left(x+3\right)}=\frac{375}{376}\)
\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{375}{376}\)
\(1-\frac{1}{x+3}=\frac{375}{376}\)
\(\frac{x+2}{x+3}=\frac{375}{376}\)
=> x + 2 = 375
=> x = 375 - 2
=> x = 373