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NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
23 tháng 7 2023
\(x\) + \(\dfrac{1}{10}\) + \(x\) + \(\dfrac{1}{11}\) = \(x\) + \(\dfrac{1}{21}\)
\(x+x-x\) = \(\dfrac{1}{21}\) - \(\dfrac{1}{10}\) - \(\dfrac{1}{11}\)
\(x\) = - \(\dfrac{331}{2310}\)
VN
Võ Ngọc Phương
VIP
23 tháng 7 2023
\(x+\dfrac{1}{10}+x+\dfrac{1}{11}=x+\dfrac{1}{21}\)
\(x+x-x=\dfrac{1}{21}-\dfrac{1}{10}-\dfrac{1}{11}\)
\(x=\dfrac{-11}{210}-\dfrac{1}{11}\)
\(x=\dfrac{-331}{2310}\)
TN
1
21 tháng 2 2023
\(\Leftrightarrow2\left(\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2010}{2012}\)
\(\Leftrightarrow\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2010}{4024}=\dfrac{1005}{2012}\)
=>1/x+1=-251/1006
=>x+1=-1006/251
=>x=-1257/251
4 tháng 1 2019
\(\left(x+1\right)\left(21-1\right)=-21\)
\(\left(x+1\right)\cdot20=-21\)
\(x+1=\frac{-21}{20}\)
\(x=\frac{-21}{20}-1\)
\(x=\frac{-41}{20}\)
4 tháng 1 2019
( x + 1 ) . ( 21 - 1 ) = -21
( x + 1 ) . 20 = -21
( x + 1 ) = -21 : 20
( x + 1 ) = 1.05
x = 1,05 - 1
x = - 0 ,05
ND
0
TT
0
K
26 tháng 3 2018
\(\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{2}{x\left(x+1\right)}=\frac{2010}{2012}\)
\(\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{x\left(x+1\right)}=\frac{2010}{2012}\)
\(\frac{2}{4.5}+\frac{2}{5.6}+\frac{2}{6.7}+...+\frac{2}{x\left(x+1\right)}=\frac{2010}{2012}\)
\(2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2010}{2012}\)
\(2\left(\frac{1}{4}-\frac{1}{x+1}\right)=\frac{2010}{2012}\)
\(\frac{1}{4}-\frac{1}{x+1}=\frac{2010}{2012}\div2\)
\(\frac{1}{4}-\frac{1}{x+1}=\frac{1005}{2012}\)
\(\frac{1}{x+1}=\frac{1}{4}-\frac{1005}{2012}\)
\(\frac{1}{x+1}=\frac{-502}{2012}=-\frac{251}{1006}\)
\(\Rightarrow x+1=1\div-\frac{251}{1006}=-\frac{1006}{251}\)
\(x=\frac{-1006}{251}-1=-\frac{1257}{251}\)
KK
9 tháng 4 2021
a) (x - 3)(y - 3) = 9 = 1.9 = 3.3
Lập bảng:
| x - 3 | 1 | -1 | 3 | -3 | 9 | -9 |
| y - 3 | 9 | -9 | 3 | -3 | 1 | -1 |
| x | 4 | 2 | 6 | 0 | 12 | -3 |
| y | 12 | -6 | 6 | 0 | 4 | 2 |
Vậy ...
KK
9 tháng 4 2021
b) A = \(\frac{10^{19}+1}{10^{20}+1}\) => 10A = \(\frac{10^{20}+10}{10^{20}+1}=1+\frac{9}{10^{20}+1}\)
B = \(\frac{10^{20}+1}{10^{21}+1}\) => 10B = \(\frac{10^{21}+10}{10^{21}+1}=1+\frac{9}{10^{21}+1}\)
Do \(10^{20}+1< 10^{21}+1\) => \(\frac{9}{10^{20}+1}>\frac{9}{10^{21}+1}\) => 10A > 10B => A > B
TT
3
S
15 tháng 5 2023
\(A=\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{120}\)
\(=2\left(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+...+\dfrac{1}{240}\right)\)
\(=2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{15.16}\right)\)
\(=2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)
\(=2\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\)
\(=2\cdot\dfrac{3}{16}\)
\(=\dfrac{3}{8}\)
\(A:x=759\)
\(\dfrac{3}{8}:x=759\)
\(\Rightarrow x=\dfrac{3}{8}:759=\dfrac{1}{2024}\)
#AvoidMe
15 tháng 5 2023
A=2(1/20+1/30+...+1/240)
=2(1/4-1/5+1/5-1/6+...+1/15-1/16)
=2*3/16=3/8
A:x=759
=>x=3/8:759=1/2024
\(x\) + \(\dfrac{1}{10}\) + \(x\) + \(\dfrac{1}{10}\) = \(x\) + \(\dfrac{1}{21}\)
\(x+x\) - \(x\) = \(\dfrac{1}{21}\) - \(\dfrac{1}{10}\) - \(\dfrac{1}{10}\)
\(x\) = \(\dfrac{1}{21}\) - \(\dfrac{1}{5}\)
\(x\) = - \(\dfrac{16}{105}\)