Tìm x:
D=\(\frac{3}{1.4}\)+\(\frac{3}{4.7}\)+...+\(\frac{1}{x.\left(x+3\right)}\)=\(\frac{100}{101}\)
Giúp mik vs nha
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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 ^_-
26 tháng 3 2020
a, \(\frac{1}{1.4}\)+\(\frac{1}{4.7}\)+......+\(\frac{1}{97.100}\)= |\(\frac{x}{3}\)|
\(\Rightarrow\)\(\frac{1}{3}\) ( \(\frac{3}{1.4}\)+\(\frac{3}{4.7}\)+.......+\(\frac{3}{97.100}\))= |\(\frac{x}{3}\)|
\(\Rightarrow\)\(\frac{1}{3}\) ( 1 - \(\frac{1}{4}\)+ \(\frac{1}{4}\)-\(\frac{1}{7}\)+......+\(\frac{1}{97}\)-\(\frac{1}{100}\)) = |\(\frac{x}{3}\)|
\(\Rightarrow\)\(\frac{1}{3}\) ( 1-\(\frac{1}{100}\)) = |\(\frac{x}{3}\)|
\(\Rightarrow\)\(\frac{1}{3}\) . \(\frac{99}{100}\) = |\(\frac{x}{3}\)|
\(\Rightarrow\)\(\frac{33}{100}\) = |\(\frac{x}{3}\)|
\(\Rightarrow\)\(\frac{x}{3}\)= \(\orbr{\begin{cases}\frac{33}{100}\\\frac{-33}{100}\end{cases}}\)
Với \(\frac{x}{3}\) = \(\frac{33}{100}\)
\(\Rightarrow\)100x= 33.3
\(\Rightarrow\)100x=99
\(\Rightarrow\)x=\(\frac{99}{100}\)
Với \(\frac{x}{3}\)=\(\frac{-33}{100}\)
\(\Rightarrow\)100x=-33.3
\(\Rightarrow\)100x=-99
\(\Rightarrow\)x=\(\frac{-99}{100}\)
Vậy x=\(\orbr{\begin{cases}\frac{99}{100}\\\frac{-99}{100}\end{cases}}\)
26 tháng 3 2020
b, \(\frac{4}{1.5}\)+ \(\frac{4}{5.9}\)+......+ \(\frac{4}{97.101}\)= |\(\frac{5x-4}{101}\)|
\(\Rightarrow\)1-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{9}\)+......+\(\frac{1}{97}\)-\(\frac{1}{101}\)= |\(\frac{5x-4}{101}\)|
\(\Rightarrow\)1-\(\frac{1}{101}\)= |\(\frac{5x-4}{101}\)
\(\Rightarrow\) \(\frac{100}{101}\)= |\(\frac{5x-4}{101}\)|
\(\Rightarrow\)\(\frac{5x-4}{101}\) =\(\orbr{\begin{cases}\frac{100}{101}\\\frac{-100}{101}\end{cases}}\)
Với \(\frac{5x-4}{101}\) =\(\frac{100}{101}\)
\(\Rightarrow\)(5x-4).101=100.101
\(\Rightarrow\)505x-404=10100
\(\Rightarrow\)505x=10504
\(\Rightarrow\)x=\(\frac{104}{5}\)
Với \(\frac{5x-4}{101}\)=\(\frac{-100}{101}\)
\(\Rightarrow\)(5x-4). 101=-100.101
\(\Rightarrow\)505x-404=-10100
\(\Rightarrow\)505x=-9696
\(\Rightarrow\)x=\(\frac{-96}{5}\)
Vậy x=\(\orbr{\begin{cases}\frac{104}{5}\\\frac{-96}{5}\end{cases}}\)
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
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:
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}=\frac{125}{376}\)
\(\Rightarrow\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x.\left(x+3\right)}\right)=\frac{125}{376}\)
\(\Rightarrow\frac{1}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{125}{376}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{x+3}=\frac{125}{376}:\frac{1}{3}=\frac{375}{376}\)
\(\Rightarrow\frac{1}{x+3}=1-\frac{375}{376}=\frac{1}{376}\Leftrightarrow x+3=376\Leftrightarrow x=373\)
7 tháng 9 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
NN
10 tháng 11 2016
Đặt vế trái phương trình là A
\(3A=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x\left(x+3\right)}\)
\(3A=\frac{4-1}{1.4}+\frac{7-4}{4.7}+\frac{10-7}{7.10}+\frac{\left(x+3\right)-x}{x\left(x+3\right)}\)
\(3A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}\)
\(3A=1-\frac{1}{x+3}=\frac{x+2}{x+3}\Rightarrow A=\frac{x+2}{3\left(x+3\right)}\)
\(\Rightarrow\frac{x+2}{3\left(x+3\right)}=\frac{667}{2002}\Rightarrow2002\left(x+2\right)=3.667.\left(x+3\right)\)
\(\Leftrightarrow2002x+4004=2001x+6003\Leftrightarrow x=1999\)
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
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
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
\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{x.\left(x+3\right)}=\frac{100}{101}\)
\(\Leftrightarrow1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{x}-\frac{1}{\left(x+3\right)}=\frac{100}{101}\)
\(\Rightarrow1-\frac{1}{x+3}=\frac{100}{101}\)
\(\Rightarrow\frac{1}{x+3}=1-\frac{100}{101}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{101}\)
\(\Rightarrow x+3=101\)
\(=>x=98\)
\(D=\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{x.\left(x+3\right)}=\frac{100}{101}\)
\(D=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{100}{101}\)
\(D=1-\frac{1}{x+3}=\frac{100}{101}\)
\(D=\frac{1}{x+3}=1-\frac{100}{101}\)
\(D=\frac{1}{x+3}=\frac{1}{101}\)
\(\Rightarrow x+3=101\Rightarrow x=98\)
Ủng hộ mk nha ^_^