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11 tháng 7 2016
\(A=\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{x\left(x+3\right)}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+3}\)
\(=1-\frac{1}{x+3}\)
\(=\frac{x+2}{x+3}=\frac{100}{101}\)
\(\Rightarrow x=98\)
11 tháng 7 2016
\(A=\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{x.\left(x+3\right)}=\frac{100}{101}\)
\(A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{100}{101}\)
\(A=1-\frac{1}{x+3}=\frac{100}{101}\)
\(\frac{1}{x+3}=1-\frac{100}{101}=\frac{1}{101}\)
=> x + 3 = 101
=> x = 101 - 3
=> x = 98
Vậy x = 98
Ủng hộ mk nha ^_-
11 tháng 7 2016
\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{x.\left(x+3\right)}=\frac{100}{101}\)
\(\Leftrightarrow1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{x}-\frac{1}{\left(x+3\right)}=\frac{100}{101}\)
\(\Rightarrow1-\frac{1}{x+3}=\frac{100}{101}\)
\(\Rightarrow\frac{1}{x+3}=1-\frac{100}{101}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{101}\)
\(\Rightarrow x+3=101\)
\(=>x=98\)
VA
11 tháng 7 2016
\(D=\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{x.\left(x+3\right)}=\frac{100}{101}\)
\(D=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{100}{101}\)
\(D=1-\frac{1}{x+3}=\frac{100}{101}\)
\(D=\frac{1}{x+3}=1-\frac{100}{101}\)
\(D=\frac{1}{x+3}=\frac{1}{101}\)
\(\Rightarrow x+3=101\Rightarrow x=98\)
Ủng hộ mk nha ^_^
NH
1
LH
16 tháng 7 2015
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x.\left(x+3\right)}=\frac{667}{2002}\)
\(=\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x.\left(x+3\right)}\right)=\frac{667}{2002}\)
\(=\frac{1}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{667}{2002}\)
\(=\frac{1}{3}.\left(\frac{1}{1}-\frac{1}{x+3}\right)=\frac{667}{2002}\)
\(\frac{1}{1}-\frac{1}{x+3}=\frac{667}{2002}:\frac{1}{3}\)
\(\frac{1}{1}-\frac{1}{x+3}=\frac{2001}{2002}\)
\(\frac{1}{x+3}=1-\frac{2001}{2002}\)
\(\frac{1}{x+3}=\frac{1}{2002}\)
\(\frac{1}{x}=\frac{1}{2002-3}\)
\(\frac{1}{x}=\frac{1}{1999}\)
Vậy x = 1999
TN
14 tháng 5 2016
đặt VT là A ta có:
\(3A=3\left(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{x\left(x+3\right)}\right)\)
\(3A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{6}{19}\)
\(3A=1-\frac{1}{x+3}\)
\(A=\left(1-\frac{1}{x+3}\right):3\)
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
DY
3
27 tháng 4 2019
1/ 1.4+ 1/ 4.7+ 1/ 7.10+....+1/ x.( x+ 3)= 672/ 2017
(3/1.4+3/4.7+3/7.10+...+ 3/x(x+3)).1/3=672/2017
(1/1-1/4+1/4-1/7+1/7-1/10+.....+(x+3)-x/x.(x+3)).1/3=672/2017
(1/1-1/(x+3)).1/3=672/2017
1/1-1/(x+3)= 672/2017:1/3
1/1-1/(x+3) = 2016/2017
1/(x+3)=1/1-2016/2017
1/(x+3)=1/2017
x+3=2017
x= 2017-3
x= 2014
MIK CHẮC CHẮN 100% LÀ ĐÚNG, DẠNG TOÁN NÀY MIK LM NHIỀU R
HOK TỐT
27 tháng 4 2019
\(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+...+\frac{1}{x\cdot\left(x+3\right)}=\frac{672}{2017}\)
\(\Rightarrow\frac{1}{3}\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{x\cdot\left(x+3\right)}\right)=\frac{672}{2017}\)
\(\Rightarrow\frac{1}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{672}{2017}\)
\(\Rightarrow\frac{1}{3}\cdot\left(1-\frac{1}{x+3}\right)=\frac{672}{2017}\Rightarrow1-\frac{1}{x+3}=\frac{672}{2017}:\frac{1}{3}\)
\(\Rightarrow1-\frac{1}{x+3}=\frac{672}{2017}\cdot3=\frac{2016}{2017}\Rightarrow\frac{1}{x+3}=1-\frac{2016}{2017}\)
\(\Rightarrow\frac{1}{x+3}=\frac{2017}{2017}-\frac{2016}{2017}\Rightarrow\frac{1}{x+3}=\frac{1}{2017}\)
\(\Rightarrow x+3=2017\Rightarrow x=2017-3\Rightarrow x=2014\)
28 tháng 8 2023
\(\dfrac{3}{1\times4}x+\dfrac{3}{4\times7}x+\dfrac{3}{7\times10}x+...+\dfrac{3}{31\times34}x=33\)
\(x\left(\dfrac{3}{1\times4}+\dfrac{3}{4\times7}+\dfrac{3}{7\times10}+...+\dfrac{3}{31\times34}\right)=33\)
\(x\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{31}-\dfrac{1}{34}\right)=33\)
\(x\left(1-\dfrac{1}{34}\right)=33\)
\(\dfrac{33}{34}x=33\)
\(x=34\)
VN
Võ Ngọc Phương
VIP
28 tháng 8 2023
\(\dfrac{3}{1.4}x+\dfrac{3}{4.7}x+\dfrac{3}{7.10}x+...+\dfrac{3}{31.34}x=33\)
\(x.3\left(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{31.34}\right)=33\)
\(x.3.\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{31}-\dfrac{1}{34}\right)=33\)
\(x.\left(1-\dfrac{1}{34}\right)=33\)
\(x.\dfrac{33}{34}=33\)
\(x=33:\dfrac{33}{34}=33.\dfrac{34}{33}\)
\(x=34\)
Đoạn cuối đáng là \(\frac{3}{x.\left(x+3\right)}\) nhưng bạn ghi lộn nha!
\(\Rightarrow1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{x}-\frac{1}{x+3}=\frac{100}{101}\)
\(\Rightarrow1-\frac{1}{x+3}=\frac{100}{101}\)
\(\Rightarrow\frac{x+2}{x+3}=\frac{100}{101}\Rightarrow x=100-2\)
\(\Rightarrow x=98\)
\(\frac{3}{1.4}+\frac{3}{4.7}+......+\frac{1}{x.\left(x+3\right)}=\frac{100}{101}\)
\(\Rightarrow1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+........+\frac{1}{x}-\frac{1}{x+3}=\frac{100}{101}\)
\(\Rightarrow1-\frac{1}{x+3}=\frac{100}{101}\)
\(\Rightarrow\frac{1}{x+3}=1-\frac{100}{101}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{101}\)
\(\Rightarrow x+3=101\)
\(\Rightarrow x=98\)